Solutions Manual to Accompany Differential Equations for ...

Solutions Manual to Accompany Differential Equations for Engineers and Scientists

By Y. Cengel and W. Palm III

Solutions to Problems in Chapter Four

Problems 4-75 Through 4-126

Solutions prepared by:

Tahsin Engin, University of Sakarya

And William Palm III, University of Rhode Island

© Solutions Manual Copyright 2012 The McGraw-Hill Companies. All rights reserved. No part of this manual may be displayed, reproduced, or distributed in any form or by any means without the written permission of the publisher or used beyond the limited distribution to teachers or educators permitted by McGraw-Hill for their individual course preparation. Any other reproduction or translation of this work is unlawful. This work is only for non-profit use by instructors in courses for which the textbook has been adopted. Any other use without publisher's consent is unlawful.

4-7 Nonhomogeneous Equation: The Method of Variation of Parameters

4-75C An th order of differential equation requires to be handled Wronskian determinant, and ( ) ( ) determinants for the functions to be determined. As the dimensions of a determinant increases, its elaboration becomes highly complicated; this is the major concern when using the method of variation of parameters to solve three or higher order differential equations.

In Problems 4-76 to4-82, we are to determine the particular solution of the given nonhomogeneous equation.

(a) Given: ( ) Solution: The related homogeneous equation and its characteristic equation are

( ) ( )( )

The roots of this equation are and . Thus the solution of the homogeneous

equation is

Before we apply the method of variation of parameters, we determine the Wronskian

( ) and the functions and . Taking ,

and we compute

( ) ( )( ) |

The functions , and are determined by substituting these expressions into Eq. 4-39,

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

Thus the particular solution of the given nonhomogeneous equation is given by

( ) ) ( ) (

( )) ( )

or simplifying

Noting that

is already a solution to the related homogeneous equation, it will be

combined with the term , leaving the particular solution in its simplest form of

Finding particular solution using the method of undetermined coefficients:

The general form of a particular solution to the nonhomogeneous term is

( ) ( )

Since any constant multiple of is a solution to the related homogeneous equation.

Differentiating yields

( ) ( )

Substituting these expressions into the nonhomogeneous equation and simplifying, we obtain

Equating coefficient, we see that and . Thus the particular solution is

given by

Maple solution:

> restart; > a := 1; b := 0; c := 0; d := 0; g := -16: > R := x*exp(2*x); > ode := a*(diff(diff(diff(diff(y(x), x), x), x), x))+b*(diff(diff(diff(y(x), x), x), x))+c*(diff(diff(y(x), x), x))+d*(diff(y(x), x))+g*y(x) = R;

> F := R = lhs(ode);

> CE := a*m^4+b*m^3+c*m^2+d*m+g = 0;

> CE := factor(%);

> Root := solve(CE);

> HomogeneousPart := lhs(subs(R = 0, ode));

> y[h] := rhs(dsolve(HomogeneousPart));

> yp := (A*x^2+B*x)*exp(2*x);

> diff(yp(x), x);

> diff(yp(x), x, x);

> diff(yp(x), x, x, x);

> diff(yp(x), x, x, x, x);

> F1 := eval(simplify(subs(y(x) = yp, F)));

> collect(collect(simplify(collect(F1, x)/exp(2*x)), x^2), x);

> eqns := [coeffs(rhs(%), x)];

> systemofequations := {eqns[1] = 0, eqns[2] = 1};

> Coefficients := solve(systemofequations, {A, B});

> y[p] := subs(Coefficients, yp);

> y = y[h]+y[p];

(b) Given: ( )

Solution: The related homogeneous equation and its characteristic equation are

( ) ( )( )

equation is

( ) and the functions and . Taking ,

and we compute

( ) ( )( ) |

The functions , and are determined by substituting these expressions into Eq. 4-39,

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

) ( ) (

| |) ( ) (

or rearranging

4-77 (a) Given: Solution: The related homogeneous equation and its characteristic equation are

equation is

( ) and the functions and . Taking , and , we

compute

( ) ( ) |

The functions and are determined by substituting these expressions into Eq. 4-39,

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

where we have used product-to-sum trigonometric identities. Thus the particular solution of the

given nonhomogeneous equation is given by

) ( ) (

or simplifying

Substituting these expressions into the nonhomogeneous equation, we obtain

Equating coefficient, we see that . Thus the particular solution is given by

(b) Given:

equation is

( ) and the functions ( ) ( ) and ( ) . Taking , and

, we compute

( ) ( ) |

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

(√ )

∫ ( ) ( )

( ) ∫

( )( )

(√ )

( )) ( ) (

(√ )) ( )

(√ ))

equation is

compute

( ) ( ) |

∫ ( ) ( )

( ) ∫

( )( ( ))

∫ ( )

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

) ) ( )

) ( ) (

or simplifying

( ) ( )

Equating coefficient, we see that and . Thus the particular solution is

given by

Maple solution:

> restart; > a := 1; b := -2; c := 0; d := 0; > R := exp(2*x)*cos(3*x); > ode := a*(diff(diff(diff(y(x), x), x), x))+b*(diff(diff(y(x), x), x))+c*(diff(y(x), x))+d*y(x) = R;

> F := R = lhs(ode);

> CE := a*m^3+b*m^2+c*m+d = 0;

> CE := factor(%);

> Root := solve(CE);

> HomogeneousPart := lhs(subs(R = 0, ode));

> y[h] := rhs(dsolve(HomogeneousPart));

> yp := A*exp(2*x)*cos(3*x)+B*exp(2*x)*sin(3*x);

> diff(yp, x);

> collect(collect(%, sin(3*x)), cos(3*x));

> diff(yp, x, x);

> diff(yp, x, x, x);

> F1 := eval(simplify(subs(y(x) = yp, F)));

> simplify(collect(F1, x)/exp(2*x));

> eqns := [coeffs(rhs(%), cos(3*x))];

> systemofequations := {eqns[1] = 0, eqns[2] = 1};

> Coefficients := solve(systemofequations, {A, B});

> y[p] := subs(Coefficients, yp);

> y = y[h]+y[p];

(b) Given: Solution: The related homogeneous equation and its characteristic equation are

equation is

compute

( ) ( ) |

∫ ( ) ( )

( ) ∫

( )( ( ))

∫( )

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

∫( ) ) ( ) (

( )) ( ) (

∫ ) ( )

or simplifying

∫[( ) ]

equation is

compute

( ) ( ) |

∫ ( ) ( )

( ) ∫

( )( ))

∫ ( ) ( )

( ) ∫

( )( )

∫( )( )

( ) ( )

∫ ( ) ( )

( ) ∫

( )( )

∫( )

( ) ( )

) ( ) (( ) ( ) )( )

(( ) ( ) )( )

or simplifying

Since any real constant is already a solution to the related homogeneous equation, we can

express the particular solution as

Equating coefficient, we see that , , and Thus the particular

solution is given by

(b) Given:

equation is

compute

( ) ( ) |

∫ ( ) ( )

( ) ∫

( )( ))

∫ ( ) ( )

( ) ∫

( ) ( )

∫ ( ) ( )

( ) ∫

( ) ( )

( | |)( ) ( ∫

) ( ) ( ∫

or simplifying

| | ∫

( )( )

equation is

( ) and the functions and . Taking , and

, we compute

( ) ( )

( ) ( ) |

( ( )|

( ) ( ) |

| | ( )

( ) ( ) |

| | ( )

∫ ( ) ( )

( ) ∫

( )( ))

∫( )

∫ ( ) ( )

∫( )( )

∫ ( ) ( )

∫( )( )

( ) ) ( )

( )) ( )

or simplifying

Equating coefficient, we see that and Thus the particular solution is

given by

(b) Given: Solution: The related homogeneous equation and its characteristic equation are

( )( )

equation is

, we compute

( ) ( )

( ) ( ) |

( ( )|

( ) ( ) |

| | ( )

( ) ( ) |

| | ( )

∫ ( ) ( )

( ) ∫

( )( ))

∫ ( ) ( )

∫( )( )

∫ ( ) ( )

∫( )( )

∫ ) ( ) (

∫( )( ) ) ( )

∫( )( )

The roots of this equation are . Thus the solution of the homogeneous equation is

( ) and the functions and . Taking , and , we compute

( ) ( ) |

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

∫ ( )

∫ ( ) ( )

( ) ∫

( )( )

( ) ) ( ) ( ( ) )( )

( ) ) ( )

or simplifying

( ) ( )

Equating coefficients, we see that and Thus the particular solution is

given by

( ) ( )

(b) Given:

( ) ( ) |

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

( ) ( )

∫ ( ) ( )

( ) ∫

( ) ( )

) ( ) ( | |)( ) (

or simplifying

( )( )

equation is

, we compute

( ) ( )

( ) ( ) |

( ) ( )|

( ) ( ) |

| | ( )

( ) ( ) |

| | ( )

We can split the non homogeneous term ( ) into two parts for convenience. Setting

( ) the functions and associated with ( ) are determined by

substituting these expressions into Eq. 4-39,

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

∫( )

∫ ( ) ( )

( ) ∫

( )( )

∫( )

Thus the particular solution due to ( ) is given by

) ( ) (

( ) ) ( )

or simplifying

The functions and associated with ( ) are determined in the same manner by

substituting these expressions into Eq. 4-39

∫ ( ) ( )

( ) ∫

( )( )

∫ ( ) ( )

( ) ∫

( )( )

∫( )

∫ ( ) ( )

( ) ∫

( )( )

∫( )

Thus the particular solution due to ( ) is given by

) ( ) (

( ) ) ( )

or simplifying

Noting that

is already a solution to the related homogeneous equation, it will be

combined with the term , leaving the particular solution in its simplest form of

since is a solution to the associated homogeneous equation. Differentiating yields

Equating coefficients, we see that Thus the particular solution is given by

(b) Given:

( )( )

equation is

, we compute

( ) ( )

( ) ( ) |

( ) ( )|

( ) ( ) |

| | ( )

( ) ( ) |

| | ( )

We can split the non homogeneous term ( ) into two parts for convenience. Setting

( ) the functions and associated with ( ) are determined by

substituting these expressions into Eq. 4-39,

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

) ( ) (

) ) ( )

or rearranging

4.7 The Euler Equations

4-83C Theorem 4-9 guarantees that the transformation will always convert the order

Euler equations to linear equations with constant coefficients, for which the finding of the

characteristic equations is straightforward and very easy. For higher order differential

equations, however, obtaining the transformed equation by employing the transformation

is quite lengthy and tedious. A more effective transformation can be used as a

short cut to come up with the same result. This results in an degree polynomial equation in

, whose roots .. will be used to determine the general solution of the transformed Euler

equation with constant coefficients. Therefore the transformation is much more practical

when obtaining the transformed equation.

4-84C Let be a three-fold repeated real root. Then the homogeneous solution

to the given Euler equation due to this three-fold repeated real root would be, from Fig. 4-17,

( ) [ ( ) ]

4-85C If is a triple root of the characteristic equation of a six order Euler equation, then

the general solution to the related homogeneous equation is, from Eq. 4-44,

[ ( ) ( )] [ ( ) ( )]

( ) [ ( ) ( )]

In Problems 4-86 through 4-92, we are to determine the general solution of the given Euler equation for .

4-86 Given:

Solution: Taking and substituting it and its derivatives , ( )

and ( )( ) into the given differential equation yields

( )( ) ( )

( )( )( )

since and thus cannot be zero. The roots of this equation are and .

Therefore the general solution of this Euler equation is given by

4-87 Given:

( )( ) ( )

( )( )( )

since and thus cannot be zero. The roots of this equation are and

. Therefore the general solution of this Euler equation is given by

4-88 Given:

( )( ) ( )

4-89 Given: or equivalently .

( )( )

since and thus cannot be zero. The roots of this equation are and √ .

( ) [ (√ ) (√ )]

4-90 Given:

( )( )

since and thus cannot be zero. The roots of this equation are and √ .

( ) (√ ) (√ )

4-91 Given:

( )( ) ( )

( )( )

4-92 Given:

( )( ) ( )

( )( )

√ . Therefore the general solution of this Euler equation is given by

[ (√ ) (√ )]

4.8 Computer Problems

4-93 See Table 4-1 for the commands needed to find polynomial roots using the various

software packages. The answers are:

4-94 (a) In Maple:

In MuPAD:

eqn94a:=ode([y''''(x)-y(x)=0],[y(x)]): solve(eqn94a)

(b) In Maple:

In MuPAD:

eqn94b:=ode([y'''(x)+3*y''(x)+4*y'(x)+12*y(x)=0],[y(x)]): solve(eqn94b)

4-95 (a) In Maple:

In MuPAD:

eqn95a:=ode([y'''(x)-y''(x)-4*y'(x)-6*y(x)=x+5],[y(x)]): solve(eqn95a)

(b) In Maple:

In MuPAD:

eqn95b:=ode([y'''(x)-y(x)=exp(3*x)],[y(x)]): solve(eqn95b)

4-96 (a) In Maple:

In MuPAD:

eqn96a:=ode([x^3*y'''(x)+x^2*y''(x)+4*y(x)=0],[y(x)]): solve(eqn96a)

(b) In Maple:

In MuPAD:

eqn96b:=ode([x^3*y'''(x)+4*x^2*y''(x)-6*x*y'(x)-12*y(x)=0],[y(x)]): solve(eqn96b)

4-97 In Maple:

In MuPAD:

eqn97:=ode([y'''(x)-4*y''(x)+4*y'(x)=0, y(0)=0,y'(0)=1,y''(0)=0],[y(x)]): solve(eqn97)

4-98 In Maple:

In MuPAD:

eqn98:=ode([y'''(x)-3*y''(x)+3*y'(x)-y(x)=0, y(0)=1,y'(0)=0,y''(0)=0],[y(x)]): solve(eqn98)

4-99 (a) In Maple:

In MuPAD:

w11:=exp(x):w12:=x*exp(x):w13:=x^2*exp(x): W:=matrix([[w11,w12,w13],[diff(w11,x),diff(w12,x),diff(w13,x)], [diff(diff(w11,x),x),diff(diff(w12,x),x),diff(diff(w13,x),x)]]): linalg:det(W)

(b) In Maple:

In MuPAD:

w11:=exp(x):w12:=2*exp(x):w13:=-3*x^2*exp(x): W:=matrix([[w11,w12,w13],[diff(w11,x),diff(w12,x),diff(w13,x)],[diff(diff(w11,x),x),diff(diff(w12,x),x),diff(diff(w13,x),x)]]): linalg:det(W)

4-100 (a) In Maple:

In MuPAD:

w11:=1/x:w12:=x^2:w13:=1: W:=matrix([[w11,w12,w13],[diff(w11,x),diff(w12,x),diff(w13,x)], [diff(diff(w11,x),x),diff(diff(w12,x),x),diff(diff(w13,x),x)]]): linalg:det(W)

(b) In Maple:

In MuPAD:

w11:=exp(-ln(x)):w12:=x^2:w13:=5: W:=matrix([[w11,w12,w13],[diff(w11,x),diff(w12,x),diff(w13,x)], [diff(diff(w11,x),x),diff(diff(w12,x),x),diff(diff(w13,x),x)]]): linalg:det(W)

Review Problems

4-101 Given: We are to verify that if is a solution of a third order linear homogeneous differential equation, then the substitution ( ) reduces the given differential equation to a second order linear equation in . Solution: Successive derivatives of ( ) ( ) can be obtained by the product rule of the derivative, such that

Substituting these expressions into the given third order linear homogeneous equation leads to

) ( ) which can be rearranged to give

The last term in the bracket vanishes since is a solution to the given third linear homogeneous

differential equation. Therefore the substitution ( ) reduces the given differential

equation to a second order linear equation in . Setting , the second order differential

equation is given by

which is the desired verification.

In Problems 4-102 through 4-126, we are to determine the general solution of the given linear

differential equation for . Arbitrary constants are also to be determined when the initial

conditions are specified.

4-102 Given: ( )

Solution: The corresponding characteristic equation is

which can be factored as

( )( )

The roots of this equation are and . Then the general solution to the given

differential equation is

4-103 Given:

( )( ) The roots of this equation is and . Thus the solution to the homogeneous part of this equation is

( ) ( )

( ) ( ) ( )

( ) ( )

Substituting these expressions into the nonhomogeneous equation, after some algebra, we have

[( ) ( )] [( ) ( )]

Equating the coefficients of like terms gives the following system of algebraic equations

yielding , , and . Therefore the general solution of the

nonhomogeneous equation becomes

4-104 Given:

(a) Given:

( ) ( ) |

∫ ( ) ( )

( ) ∫

( ) ( )

∫( )

∫ ( ) ( )

( ) ∫

( ) ( )

∫( )

∫ ( ) ( )

( ) ∫

( ) ( )

( )( ) (

or simplifying

Noting that

is already a solution to the related homogeneous equation, it will be combined

with the term , leaving the particular solution in its simplest form of

4-106 Given:

and ( )( ) into the related homogeneous equation yields

( )( ) ( )

( )( )

. Therefore the solution to the homogeneous part of this Euler equation is given by

Now we divide both sides of the given equation by to get an equation in standard form, that

√ , we compute

Noting that ( )

in the standard form of the equation, the functions and are

determined by substituting these expressions into Eq. 4-39,

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

( ) ( )

∫( )

( )√

) ( ) (

( )√ ) (

Finally, the general solution of the differential equation is obtained by combining the

homogeneous solution with this particular solution:

where .

4-107 Given:

( )( )

The roots of this equation is and √ . Thus the solution to the

homogeneous part of this equation is

( √ √ )

Substituting these expressions into the nonhomogeneous equation, we have

( ) ( ) ( )

or simplifying

( ) ( )

Equating the coefficients of like terms yields , , and .

Therefore the general solution of the nonhomogeneous equation becomes

( ) ( √ √ )

4-108 Given: ( )

The roots of this equation are . Then the general solution to the given

4-109 Given:

and ( )( ) into the related homogeneous equation yields

( )( ) ( )

Therefore the solution to the homogeneous part of this Euler equation is given by

Now we divide both sides of the given equation by to get an equation in standard form, that

compute

Noting that ( )

in the standard form of the equation, the functions and are

determined by substituting these expressions into Eq. 4-39,

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

∫ ( ) ( )

( ) ∫

) ( ) (

Finally, the general solution of the differential equation is obtained by combining the

homogeneous solution with this particular solution:

where .

4-110 Given:

4-111 Given:

The roots of this equation are . Then the general solution to the given differential

equation is

4-112 Given: , ( ) , ( ) ( )

Differentiating, we obtain

Hence, the initial conditions imply that

or simplifying,

from which , and . Substituting these constants into the general solution,

we obtain

( )( )

The roots of this equation are and

. Thus the solution to the homogeneous

part of this equation is

The general form of a particular solution to the nonhomogeneous term .

Equating coefficients, we see that , and . Thus a general solution to the

nonhomogeneous differential equation is given by

( ) The roots of this equation are and . Thus the solution to the homogeneous part

of this equation is

The general form of a particular solution to the nonhomogeneous term

Equating coefficients, we see that and . Thus a general solution to the

4-115 Given:

( )( )

The roots of this equation are and

. Thus the solution to the

The general form of a particular solution to the nonhomogeneous term .

( ) ( )

or simplifying

Equating coefficients, we see that , and . Thus a general solution to the

4-116 Given: ( )

( ) The roots of this equation are and . Thus the solution to the homogeneous

The general form of a particular solution to the nonhomogeneous term ( ) is

since a linear function of is a solution of the related homogeneous equation. Differentiating

yields

or simplifying

Equating coefficients, we see that , and . Thus the particular

solution due to the nonhomogeneous term is given by

The general form of a particular solution to the nonhomogeneous term ( ) is

Differentiating and combining like terms yields

( ) ( )

or simplifying

( ) ( )

Equating coefficients, we see that and . Thus the particular solution due to

the nonhomogeneous term is given by

Then the general solution to the given nonhomogeneous differential equation can be expressed

4-117 Given: ( )

The roots of this equation is

√ and

√ . Thus the solution to the

√ ) √ (

( ) ( )

[ ] [ ]

Therefore the general solution of the nonhomogeneous equation becomes

√ ) √ (

4-118 Given:

( )( )

The roots of this equation is and

. Thus the solution to the

( ) ( )

( ) ( ) [( ) ( ) ]

[( ) ]

or simplifying

Equating the coefficients of like terms yields , and . Therefore the

general solution of the nonhomogeneous equation becomes

( ) √ ( √ √ ) √ ( √ √ )

4-119 Given: ( ) Solution: The corresponding characteristic equation is

The roots of this equation is √ √ and √ √ . Thus the solution to the homogeneous part of this equation is

√ ( √ √ ) √ ( √ √ )

( ) [( ) ]

or simplifying

Equating the coefficients of like terms yields , and . Therefore

the general solution of the nonhomogeneous equation becomes

√ ( √ √ ) √ ( √ √ )

4-120 Given: ( ) , ( ) , ( ) ( ) ( )

( )( )

Differentiating, we obtain

Hence, the initial conditions imply that

or simplifying,

from which , , and . Substituting these constants into the

general solution, we obtain

Maple solution: > restart; > with(DEtools); > ode := diff(y, x, x, x, x)-y = 0;

> y := rhs(dsolve(ode));

> y1 := diff(y, x);

> y2 := diff(y, x, x);

> y3 := diff(y, x, x, x);

> Eq1 := eval(subs(x = 0, y)) = 1;

> Eq2 := eval(subs(x = 0, y1)) = 0;

> Eq3 := eval(subs(x = 0, y2)) = 0;

> Eq4 := eval(subs(x = 0, y3)) = 1;

> solC := solve({Eq1, Eq2, Eq3, Eq4}, {_C1, _C2, _C3, _C4});

> Y := subs(solC, y);

4-121 Given:

( ) The roots of this equation is and . Thus the solution to the homogeneous part

of this equation is

( ) ( )

[( ) ( ) ( )]

from which we get , , and . Therefore the

general solution to the nonhomogeneous equation is given by

4-122 Given:

( )( ) ( )

( )( )

. Therefore the general solution of this Euler equation is given by

4-123 ( ) ( ) ( ) ( ) ( )

( )( )

The roots of this equation are and . Then the solution of the

homogeneous part of the given differential equation is

( ) ( )

It is evident from the nonhomogeneous equation that the particular solution corresponding the

term ( ) is simply . Direct substitution gives

( )( ) ( )

Thus the general solution is given by

( ) ( ) ( )

To satisfy the initial conditions, we obtain

Substituting the initial conditions ( ) ( ) ( ) ( ) , we get the system

from which , . Substituting these constants into the general

solution, we obtain

or simply

4-124 Given: ( )

Solution: Taking and substituting it and its derivatives , ( ) ,

( )( ) and ( ) ( )( )( ) into the given differential

equation yields

( )( )( ) ( )

( )( )

( ) [ ( ) ( )] √

4-125 Given:

( )( )

[ (√ ) (√ )]

4-126 Given: , ( ) , ( ) , ( )

( ) The roots of this equation are and . Thus the solution to the homogeneous

since any constant or a linear function of is a solution of the homogeneous equation.

which can be simplified to

Equating coefficients of like terms, we obtain

yielding and . Thus a general solution to the nonhomogeneous

differential equation is given by

To satisfy the initial conditions, we obtain

Substituting the initial conditions ( ) ( ) , and ( ) we get the system

from which and . Substituting these constants into the general

solution, we obtain

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