Theory of Linear Ordinary Differential Equations · Theory of Linear Ordinary Differential...

136
logo1 Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem Theory of Linear Ordinary Differential Equations Bernd Schr ¨ oder Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Theory of Linear Ordinary Differential Equations

Transcript of Theory of Linear Ordinary Differential Equations · Theory of Linear Ordinary Differential...

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Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Theory of Linear Ordinary DifferentialEquations

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

DefinitionA linear n-th order differential equation is of the form

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x),

with an not being the constant function 0.

I It is called homogeneous if and only if g = 0.I It is called inhomogeneous if and only if g 6= 0.I If, in an inhomogeneous equation, we replace the right side

g with 0, we obtain the corresponding homogeneousequation.

Note that the coefficients are functions. The results in thispresentation apply to constant coefficient equations as well asCauchy-Euler equations or the equations that are being solvedwith series solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

DefinitionA linear n-th order differential equation is of the form

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x),

with an not being the constant function 0.I It is called homogeneous if and only if g = 0.

I It is called inhomogeneous if and only if g 6= 0.I If, in an inhomogeneous equation, we replace the right side

g with 0, we obtain the corresponding homogeneousequation.

Note that the coefficients are functions. The results in thispresentation apply to constant coefficient equations as well asCauchy-Euler equations or the equations that are being solvedwith series solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

DefinitionA linear n-th order differential equation is of the form

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x),

with an not being the constant function 0.I It is called homogeneous if and only if g = 0.I It is called inhomogeneous if and only if g 6= 0.

I If, in an inhomogeneous equation, we replace the right sideg with 0, we obtain the corresponding homogeneousequation.

Note that the coefficients are functions. The results in thispresentation apply to constant coefficient equations as well asCauchy-Euler equations or the equations that are being solvedwith series solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

DefinitionA linear n-th order differential equation is of the form

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x),

with an not being the constant function 0.I It is called homogeneous if and only if g = 0.I It is called inhomogeneous if and only if g 6= 0.I If, in an inhomogeneous equation, we replace the right side

g with 0, we obtain the corresponding homogeneousequation.

Note that the coefficients are functions. The results in thispresentation apply to constant coefficient equations as well asCauchy-Euler equations or the equations that are being solvedwith series solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

DefinitionA linear n-th order differential equation is of the form

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x),

with an not being the constant function 0.I It is called homogeneous if and only if g = 0.I It is called inhomogeneous if and only if g 6= 0.I If, in an inhomogeneous equation, we replace the right side

g with 0, we obtain the corresponding homogeneousequation.

Note that the coefficients are functions. The results in thispresentation apply to constant coefficient equations as well asCauchy-Euler equations or the equations that are being solvedwith series solutions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Existence and Uniqueness

Every initial value problem of the form

an(x)y(n)(x)+ · · ·+a1(x)y′(x)+a0(x)y(x) = g(x),y(x0) = y0,

y′(x0) = y1,...

y(n−1)(x0) = yn−1,

where an is not the constant function 0 and all ai(x) and g(x)have continuous first derivatives, has a unique solution.So, in some ways, the solutions look like n-dimensional space.We are interested in using this analogy.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Existence and UniquenessEvery initial value problem of the form

an(x)y(n)(x)+ · · ·+a1(x)y′(x)+a0(x)y(x) = g(x),y(x0) = y0,

y′(x0) = y1,...

y(n−1)(x0) = yn−1,

where an is not the constant function 0 and all ai(x) and g(x)have continuous first derivatives, has a unique solution.

So, in some ways, the solutions look like n-dimensional space.We are interested in using this analogy.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Existence and UniquenessEvery initial value problem of the form

an(x)y(n)(x)+ · · ·+a1(x)y′(x)+a0(x)y(x) = g(x),y(x0) = y0,

y′(x0) = y1,...

y(n−1)(x0) = yn−1,

where an is not the constant function 0 and all ai(x) and g(x)have continuous first derivatives, has a unique solution.So, in some ways, the solutions look like n-dimensional space.We are interested in using this analogy.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Existence and Uniqueness Theorem

, p.510

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Existence and Uniqueness Theorem

, p.510Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

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Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Furthering the Analogy Between Vectors andSolutions

Superposition Principle. Let y1 and y2 be solutions of thehomogeneous linear differential equation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= 0,

and let c1 and c2 be real numbers. Theny(x) := c1y1(x)+ c2y2(x) is a solution, too.

So solutions of homogeneous equations have the same algebraicproperties as vectors.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Furthering the Analogy Between Vectors andSolutions

Superposition Principle. Let y1 and y2 be solutions of thehomogeneous linear differential equation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= 0,

and let c1 and c2 be real numbers. Theny(x) := c1y1(x)+ c2y2(x) is a solution, too.

So solutions of homogeneous equations have the same algebraicproperties as vectors.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Furthering the Analogy Between Vectors andSolutions

Superposition Principle. Let y1 and y2 be solutions of thehomogeneous linear differential equation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= 0,

and let c1 and c2 be real numbers. Theny(x) := c1y1(x)+ c2y2(x) is a solution, too.

So solutions of homogeneous equations have the same algebraicproperties as vectors.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0+

(an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x)

)= 0

an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0+

(an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x)

)= 0

an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0+

(an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x)

)= 0

an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0+

(an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x)

)= 0

an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0+

(an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x)

)= 0

an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0

+(

an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x))

= 0an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0+

(an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x)

)= 0

an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof of the Superposition Principle

an(x)y(n)1 (x)+ · · ·+a1(x)y′1(x)+a0(x)y1(x) = 0

an(x)y(n)2 (x)+ · · ·+a1(x)y′2(x)+a0(x)y2(x) = 0

c1(

an(x)y(n)1 (x) + · · · + a0(x)y1(x)

)= c1 ·0

+c2(

an(x)y(n)2 (x) + · · · + a0(x)y2(x)

)= c2 ·0

an(x)(c1y1)(n)(x) + · · · + a0(x)(c1y1)(x) = 0+

(an(x)(c2y2)(n)(x) + · · · + a0(x)(c2y2)(x)

)= 0

an(x)(c1y1 + c2y2)(n)(x) + · · · + a0(x)(c1y1 + c2y2)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Handling Inhomogeneous Equations

For the linear inhomogeneous differential equation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x)

let yh(x) denote the general solution of the correspondinghomogeneous equation. Moreover let yp(x) be one particularsolution of the inhomogeneous equation. Then the generalsolution of the inhomogeneous equation is

y(x) = yp(x)+ yh(x).

So the theory of inhomogeneous equations is pretty muchreduced to that of homogeneous equations.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Handling Inhomogeneous EquationsFor the linear inhomogeneous differential equation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x)

let yh(x) denote the general solution of the correspondinghomogeneous equation. Moreover let yp(x) be one particularsolution of the inhomogeneous equation. Then the generalsolution of the inhomogeneous equation is

y(x) = yp(x)+ yh(x).

So the theory of inhomogeneous equations is pretty muchreduced to that of homogeneous equations.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Handling Inhomogeneous EquationsFor the linear inhomogeneous differential equation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= g(x)

let yh(x) denote the general solution of the correspondinghomogeneous equation. Moreover let yp(x) be one particularsolution of the inhomogeneous equation. Then the generalsolution of the inhomogeneous equation is

y(x) = yp(x)+ yh(x).

So the theory of inhomogeneous equations is pretty muchreduced to that of homogeneous equations.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)

an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)

an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Proof

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

+(

an(x)y(n)h (x) + · · · + a0(x)yh(x) = 0

)an(x)(yp + yh)(n)(x) + · · · + a0(x)(yp + yh)(x) = g(x)

and

an(x)y(n)i (x) + · · · + a0(x)yi(x) = g(x)

−(

an(x)y(n)p (x) + · · · + a0(x)yp(x) = g(x)

)an(x)(yi − yp)(n)(x) + · · · + a0(x)(yi − yp)(x) = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of Vectors

How do we actually know that several vectors “point indifferent directions”?

Let~v1,~v2, . . . ,~vn be vectors. Then any sumn

∑i=1

ci~vi = c1~v1 + c2~v2 + · · ·+ cn~vn

with the ci being real numbers is called a linear combinationof the vectors.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

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Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of VectorsHow do we actually know that several vectors “point indifferent directions”?

Let~v1,~v2, . . . ,~vn be vectors. Then any sumn

∑i=1

ci~vi = c1~v1 + c2~v2 + · · ·+ cn~vn

with the ci being real numbers is called a linear combinationof the vectors.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of VectorsHow do we actually know that several vectors “point indifferent directions”?

Let~v1,~v2, . . . ,~vn be vectors. Then any sumn

∑i=1

ci~vi = c1~v1 + c2~v2 + · · ·+ cn~vn

with the ci being real numbers is called a linear combinationof the vectors.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Independence for Vectors

A set of n vectors {~v1, · · · ,~vn} is called linearly dependent ifand only if there are numbers c1, . . . ,cn, which are not all zero,such that c1~v1 + · · ·+ cn~vn =~0, where~0 denotes the nullvector, for which all components are zero.If no such numbers exist, the set of vectors is called linearlyindependent. That is, a set of n vectors {~v1, · · · ,~vn} is calledlinearly independent if and only if the only numbers c1, · · · ,cn,

for whichn

∑i=1

ci~vi =~0 are c1 = c2 = · · · = cn = 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Independence for VectorsA set of n vectors {~v1, · · · ,~vn} is called linearly dependent ifand only if there are numbers c1, . . . ,cn, which are not all zero,such that c1~v1 + · · ·+ cn~vn =~0, where~0 denotes the nullvector, for which all components are zero.

If no such numbers exist, the set of vectors is called linearlyindependent. That is, a set of n vectors {~v1, · · · ,~vn} is calledlinearly independent if and only if the only numbers c1, · · · ,cn,

for whichn

∑i=1

ci~vi =~0 are c1 = c2 = · · · = cn = 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Independence for VectorsA set of n vectors {~v1, · · · ,~vn} is called linearly dependent ifand only if there are numbers c1, . . . ,cn, which are not all zero,such that c1~v1 + · · ·+ cn~vn =~0, where~0 denotes the nullvector, for which all components are zero.If no such numbers exist, the set of vectors is called linearlyindependent. That is, a set of n vectors {~v1, · · · ,~vn} is calledlinearly independent if and only if the only numbers c1, · · · ,cn,

for whichn

∑i=1

ci~vi =~0 are c1 = c2 = · · · = cn = 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1

113

+ c2

242

+ c3

3−14

=

000

1c1 + 2c2 + 3c3 = 01c1 + 4c2 − 1c3 = 03c1 + 2c2 + 4c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1

113

+ c2

242

+ c3

3−14

=

000

1c1 + 2c2 + 3c3 = 01c1 + 4c2 − 1c3 = 03c1 + 2c2 + 4c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1

113

+ c2

242

+ c3

3−14

=

000

1c1 + 2c2 + 3c3 = 0

1c1 + 4c2 − 1c3 = 03c1 + 2c2 + 4c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1

113

+ c2

242

+ c3

3−14

=

000

1c1 + 2c2 + 3c3 = 01c1 + 4c2 − 1c3 = 0

3c1 + 2c2 + 4c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1

113

+ c2

242

+ c3

3−14

=

000

1c1 + 2c2 + 3c3 = 01c1 + 4c2 − 1c3 = 03c1 + 2c2 + 4c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 0

2c2 − 4c3 = 0− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 0

0 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3

= c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2

= c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1

, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if the vectors (1,1,3), (2,4,2) and(3,−1,4) are linearly independent.

c1 + 2c2 + 3c3 = 0c1 + 4c2 − c3 = 0

3c1 + 2c2 + 4c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 4c2 − 5c3 = 0

c1 + 2c2 + 3c3 = 02c2 − 4c3 = 0

− 13c3 = 00 = c3 = c2 = c1, and the vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

2c1 − 1c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

2c1 − 1c2 + 3c3 = 0

−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

2c1 − 1c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0

−4c1 + 3c2 − 5c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

2c1 − 1c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0

c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0

c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3

, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2

,choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1

: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1

, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

2c1 − c2 + 3c3 = 0−2c1 + 2c2 − 2c3 = 0−4c1 + 3c2 − 5c3 = 0

2c1 − c2 + 3c3 = 0c2 + c3 = 0c2 + c3 = 0

c2 = −c3, c1 =c2 −3c3

2,

choose c3 = 1: c2 = −1, c1 = −2.

−2

2−2−4

−123

+

3−2−5

=

000

,

and the vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Why use Matrices?

113

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Why use Matrices?

113

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Why use Matrices?

113

242

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Why use Matrices?

113

2 34 −12 4

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Why use Matrices?

113

2 34 −12 4

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Matrices

Let m and n be positive integers. An m×n-matrix is arectangular array of mn numbers aij, commonly indexed andwritten as follows.

A=(ai,j) i = 1, . . . ,mj = 1, . . . ,n

=

a11 a12 · · · a1(n−1) a1na21 a22 · · · a2(n−1) a2na31 a32 · · · a3(n−1) a3n

......

a(m−1)1 a(m−1)2 · · · a(m−1)(n−1) a(m−1)nam1 am2 · · · am(n−1) amn

.

The index i is called the row index and the index j is called thecolumn index.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

MatricesLet m and n be positive integers. An m×n-matrix is arectangular array of mn numbers aij, commonly indexed andwritten as follows.

A=(ai,j) i = 1, . . . ,mj = 1, . . . ,n

=

a11 a12 · · · a1(n−1) a1na21 a22 · · · a2(n−1) a2na31 a32 · · · a3(n−1) a3n

......

a(m−1)1 a(m−1)2 · · · a(m−1)(n−1) a(m−1)nam1 am2 · · · am(n−1) amn

.

The index i is called the row index and the index j is called thecolumn index.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

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Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determinants

Let A =(

a11 a12a21 a22

)be a 2×2 matrix. Then we define the

determinant of A to be

det(A) := det(

a11 a12a21 a22

):=

∣∣∣∣(a11 a12a21 a22

)∣∣∣∣ := a11a22 −a12a21.

Let A = (aij)i,j=1,...,n be a square matrix and let Aij be the matrixobtained by erasing the ith row and the jth column. Then thedeterminant of A is defined recursively by

det(A) := |A| :=n

∑j=1

(−1)i+jaij det(Aij) =n

∑i=1

(−1)i+jaij det(Aij) ,

where the i in the first sum is an arbitrary row and the j in thesecond sum is an arbitrary column.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

DeterminantsLet A =

(a11 a12a21 a22

)be a 2×2 matrix. Then we define the

determinant of A to be

det(A) := det(

a11 a12a21 a22

):=

∣∣∣∣(a11 a12a21 a22

)∣∣∣∣ := a11a22 −a12a21.

Let A = (aij)i,j=1,...,n be a square matrix and let Aij be the matrixobtained by erasing the ith row and the jth column. Then thedeterminant of A is defined recursively by

det(A) := |A| :=n

∑j=1

(−1)i+jaij det(Aij) =n

∑i=1

(−1)i+jaij det(Aij) ,

where the i in the first sum is an arbitrary row and the j in thesecond sum is an arbitrary column.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

DeterminantsLet A =

(a11 a12a21 a22

)be a 2×2 matrix. Then we define the

determinant of A to be

det(A) := det(

a11 a12a21 a22

):=

∣∣∣∣(a11 a12a21 a22

)∣∣∣∣ := a11a22 −a12a21.

Let A = (aij)i,j=1,...,n be a square matrix and let Aij be the matrixobtained by erasing the ith row and the jth column. Then thedeterminant of A is defined recursively by

det(A) := |A| :=n

∑j=1

(−1)i+jaij det(Aij) =n

∑i=1

(−1)i+jaij det(Aij) ,

where the i in the first sum is an arbitrary row and the j in thesecond sum is an arbitrary column.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Uses of the Determinant

1. The determinant gives the n-dimensional volume of theparallelepiped spanned by the column vectors.

2. The n-dimensional vectors~v1, . . . ,~vn are linearlyindependent if and only if det(~v1, . . . ,~vn) 6= 0, where(~v1, . . . ,~vn) denotes a matrix whose columns are thevectors~vi.

3. Computation of characteristic polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Uses of the Determinant1. The determinant gives the n-dimensional volume of the

parallelepiped spanned by the column vectors.

2. The n-dimensional vectors~v1, . . . ,~vn are linearlyindependent if and only if det(~v1, . . . ,~vn) 6= 0, where(~v1, . . . ,~vn) denotes a matrix whose columns are thevectors~vi.

3. Computation of characteristic polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Uses of the Determinant1. The determinant gives the n-dimensional volume of the

parallelepiped spanned by the column vectors.2. The n-dimensional vectors~v1, . . . ,~vn are linearly

independent if and only if det(~v1, . . . ,~vn) 6= 0, where(~v1, . . . ,~vn) denotes a matrix whose columns are thevectors~vi.

3. Computation of characteristic polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Uses of the Determinant1. The determinant gives the n-dimensional volume of the

parallelepiped spanned by the column vectors.2. The n-dimensional vectors~v1, . . . ,~vn are linearly

independent if and only if det(~v1, . . . ,~vn) 6= 0, where(~v1, . . . ,~vn) denotes a matrix whose columns are thevectors~vi.

3. Computation of characteristic polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.

113

2 34 −12 4

det

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)

= 1 ·18−1 ·2+3 · (−14)= −26 6= 0

The vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)= 1 ·18

−1 ·2+3 · (−14)= −26 6= 0

The vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)= 1 ·18−1 ·2

+3 · (−14)= −26 6= 0

The vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)= 1 ·18−1 ·2+3 · (−14)

= −26 6= 0The vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)= 1 ·18−1 ·2+3 · (−14)= −26

6= 0The vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)= 1 ·18−1 ·2+3 · (−14)= −26 6= 0

The vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

113

,

242

and

3−14

are linearly

independent.113

2 34 −12 4

det

= 1 ·det(

4 −12 4

)−1 ·det

(2 32 4

)+3 ·det

(2 34 −1

)= 1 ·18−1 ·2+3 · (−14)= −26 6= 0

The vectors are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)

− (−2) ·det(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)

= 2 · (−4)− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)

− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)

+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)+(−4)(−4)

= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if

2−2−4

,

−123

and

3−2−5

are

linearly independent.

det

2 −1 3−2 2 −2−4 3 −5

= 2 ·det(

2 −23 −5

)− (−2) ·det

(−1 33 −5

)

+(−4) ·det(−1 32 −2

)= 2 · (−4)− (−2)(−4)+(−4)(−4)= 0

The vectors are linearly dependent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of Functions

We need to determine what it means that several functions“point in different directions”.Otherwise we would not be able to recognize that a family likeyc1,c2(x) = c1 sin2(x)+ c2

(1− cos(2x)

)is not the general

solution of sin(x)y′′− cos(x)y′ +2sin(x)y = 0.(The family has only one constant, because2sin2(x) = 1− cos(2x).)

Let f1, f2, . . . , fn be functions. Then any sumn

∑i=1

cifi = c1f1 + c2f2 + · · ·+ cnfn

with the ci being real numbers is called a linear combinationof the functions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of FunctionsWe need to determine what it means that several functions“point in different directions”.

Otherwise we would not be able to recognize that a family likeyc1,c2(x) = c1 sin2(x)+ c2

(1− cos(2x)

)is not the general

solution of sin(x)y′′− cos(x)y′ +2sin(x)y = 0.(The family has only one constant, because2sin2(x) = 1− cos(2x).)

Let f1, f2, . . . , fn be functions. Then any sumn

∑i=1

cifi = c1f1 + c2f2 + · · ·+ cnfn

with the ci being real numbers is called a linear combinationof the functions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of FunctionsWe need to determine what it means that several functions“point in different directions”.Otherwise we would not be able to recognize that a family likeyc1,c2(x) = c1 sin2(x)+ c2

(1− cos(2x)

)is not the general

solution of sin(x)y′′− cos(x)y′ +2sin(x)y = 0.

(The family has only one constant, because2sin2(x) = 1− cos(2x).)

Let f1, f2, . . . , fn be functions. Then any sumn

∑i=1

cifi = c1f1 + c2f2 + · · ·+ cnfn

with the ci being real numbers is called a linear combinationof the functions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of FunctionsWe need to determine what it means that several functions“point in different directions”.Otherwise we would not be able to recognize that a family likeyc1,c2(x) = c1 sin2(x)+ c2

(1− cos(2x)

)is not the general

solution of sin(x)y′′− cos(x)y′ +2sin(x)y = 0.(The family has only one constant, because2sin2(x) = 1− cos(2x).)

Let f1, f2, . . . , fn be functions. Then any sumn

∑i=1

cifi = c1f1 + c2f2 + · · ·+ cnfn

with the ci being real numbers is called a linear combinationof the functions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Combinations of FunctionsWe need to determine what it means that several functions“point in different directions”.Otherwise we would not be able to recognize that a family likeyc1,c2(x) = c1 sin2(x)+ c2

(1− cos(2x)

)is not the general

solution of sin(x)y′′− cos(x)y′ +2sin(x)y = 0.(The family has only one constant, because2sin2(x) = 1− cos(2x).)

Let f1, f2, . . . , fn be functions. Then any sumn

∑i=1

cifi = c1f1 + c2f2 + · · ·+ cnfn

with the ci being real numbers is called a linear combinationof the functions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Independence for Functions

A set of n functions {f1, . . . , fn} is called linearly dependent ifand only if there are numbers c1, . . . ,cn, which are not all zero,such that c1f1 + · · ·+ cnfn = 0. That is, c1, . . . ,cn must be suchthat for all x in the domain of f1, . . . , fn we havec1f1(x)+ · · ·+ cnfn(x) = 0.If no such numbers exist, then the set of functions is calledlinearly independent. That is, a set of n functions {f1, . . . , fn}is called linearly independent if and only if the only numbers

c1, . . . ,cn, for whichn

∑i=1

cifi = 0 are c1 = c2 = · · · = cn = 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Independence for FunctionsA set of n functions {f1, . . . , fn} is called linearly dependent ifand only if there are numbers c1, . . . ,cn, which are not all zero,such that c1f1 + · · ·+ cnfn = 0. That is, c1, . . . ,cn must be suchthat for all x in the domain of f1, . . . , fn we havec1f1(x)+ · · ·+ cnfn(x) = 0.

If no such numbers exist, then the set of functions is calledlinearly independent. That is, a set of n functions {f1, . . . , fn}is called linearly independent if and only if the only numbers

c1, . . . ,cn, for whichn

∑i=1

cifi = 0 are c1 = c2 = · · · = cn = 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Linear Independence for FunctionsA set of n functions {f1, . . . , fn} is called linearly dependent ifand only if there are numbers c1, . . . ,cn, which are not all zero,such that c1f1 + · · ·+ cnfn = 0. That is, c1, . . . ,cn must be suchthat for all x in the domain of f1, . . . , fn we havec1f1(x)+ · · ·+ cnfn(x) = 0.If no such numbers exist, then the set of functions is calledlinearly independent. That is, a set of n functions {f1, . . . , fn}is called linearly independent if and only if the only numbers

c1, . . . ,cn, for whichn

∑i=1

cifi = 0 are c1 = c2 = · · · = cn = 0.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

The Wronskian

Let f1, · · · , fn be (n−1) times differentiable functions. If theWronskian

W(f1, · · · , fn)(x) := det

f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

......

...f (n−1)1 (x) f (n−1)

2 (x) · · · f (n−1)n (x)

is not equal to zero for some value of x, then {f1, · · · , fn} is alinearly independent set of functions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

The WronskianLet f1, · · · , fn be (n−1) times differentiable functions. If theWronskian

W(f1, · · · , fn)(x) := det

f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

......

...f (n−1)1 (x) f (n−1)

2 (x) · · · f (n−1)n (x)

is not equal to zero for some value of x, then {f1, · · · , fn} is alinearly independent set of functions.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)

−1 ·det(

et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)

= t(

te2t +2e2t − te2t − e2t)

−(

te2t +2e2t − te2t)

= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)

−(

te2t +2e2t − te2t)

= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)

= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)= te2t −2e2t

6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Determine if t, et and tet are linearly independent.

det

t et tet

1 et tet + et

0 et tet +2et

= t ·det(

et tet + et

et tet +2et

)−1 ·det

(et tet

et tet +2et

)

+0 ·det(

et tet

et tet + et

)= t

(te2t +2e2t − te2t − e2t

)−

(te2t +2e2t − te2t

)= te2t −2e2t 6= 0

The functions are linearly independent.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Defining the General Solution

The general solution of a differential equation is a family offunctions so that for every initial value problem for thedifferential equation there is a unique choice of the coefficientsthat gives the solution of the initial value problem. A particularsolution of a differential equation is one specific solution.

In the theory, we typically work with initial value problems,because even this definition is a bit messy.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Defining the General SolutionThe general solution of a differential equation is a family offunctions so that for every initial value problem for thedifferential equation there is a unique choice of the coefficientsthat gives the solution of the initial value problem. A particularsolution of a differential equation is one specific solution.

In the theory, we typically work with initial value problems,because even this definition is a bit messy.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Defining the General SolutionThe general solution of a differential equation is a family offunctions so that for every initial value problem for thedifferential equation there is a unique choice of the coefficientsthat gives the solution of the initial value problem. A particularsolution of a differential equation is one specific solution.

In the theory, we typically work with initial value problems,because even this definition is a bit messy.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Solution Theorem for Linear HomogeneousDifferential Equations

The general solution of a linear homogeneous differentialequation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= 0,

is of the form

y(x) = c1y1(x)+ · · ·+ cnyn(x),

where {y1, · · · ,yn} is a linearly independent set of particularsolutions of the linear homogeneous differential equation.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations

logo1

Existence and Uniqueness Linear Independence Matrices and Determinants Linear Independence Revisited Solution Theorem

Solution Theorem for Linear HomogeneousDifferential Equations

The general solution of a linear homogeneous differentialequation

an(x)y(n)(x)+an−1(x)y(n−1)(x)+· · ·+a1(x)y′(x)+a0(x)y(x)= 0,

is of the form

y(x) = c1y1(x)+ · · ·+ cnyn(x),

where {y1, · · · ,yn} is a linearly independent set of particularsolutions of the linear homogeneous differential equation.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Theory of Linear Ordinary Differential Equations