The Laplace Transform - Kennesaw State...

The Laplace TransformSolving Initial Value Problems

Philippe B. Laval

KSU

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Philippe B. Laval (KSU) Solving IVP Today 1 / 10

Outline

Description of the method

Examples


Description of the Technique

In the introduction of the previous section, we used an example to showhow the Laplace transform could be used to solve initial value problems.We summarize the steps involved in this technique, then look at someexamples. In this outline, we assume that in the equation we are solving,the independent variable is t, the dependent variable is y = y (t) and wedenote the Laplace transform of y by Y (s) = L{y (t)} (s). In otherwords, y (t) = L−1 {Y (s)} (t).Generally speaking, this technique can be seen as a change of variable.Given an equation where the unknown function is y (t), using the Laplacetransform, we rewrite it as an equation in Y (s), the Laplace transform ofy (t). In other words, Y (s) = L{y (t)} (s). If we can solve this newequation for Y (s), then we can find y (t) since y (t) = L−1 {Y (s)} (t).



To solve an initial value problem involving an unknown function y (t),using the Laplace transform:

1 Take the Laplace transform on both sides of the equation.2 Using the properties of the Laplace transform and the initialconditions, obtain an equation in Y (s).

3 Solve this equation for Y (s).4 Find y (t) = L−1 {Y (s)} (t) using the techniques described in theprevious section.



RemarkOne of the formulas that will be used by this technique is the formula ofthe Laplace transform of the derivatives of a function, L

{f (n)

}(s) =

snL{f } (s)− sn−1f (0)− sn−2f ′ (0)− sn−3f ′′ (0)− ...− f (n−1) (0). Notethat this formula uses the value of the function in question and itsderivatives at t = 0. This implies that the initial conditions of our initialvalue problem must be at t = 0. We will show how to handle the casewhen the initial condition is not at t = 0 in one of the examples.


Examples

Example

Solve the initial value problem

y ′′ − 2y ′ + 5y = −8e−t

y (0) = 2y ′ (0) = 12

1 First, find Y = L{y}

2 We have Y (s) =2s2 + 10s

(s + 1) (s2 − 2s + 5)3 Earlier, we found that

L−1{

2s2 + 10s(s2 − 2s + 5) (s + 1)

}= 3et cos 2t + 4et sin 2t − e−t = y (t)


Examples

Example


y ′′ − 2y ′ + 5y = −8e−t

y (0) = 2y ′ (0) = 12


2 We have Y (s) =2s2 + 10s

(s + 1) (s2 − 2s + 5)

3 Earlier, we found that

L−1{

2s2 + 10s(s2 − 2s + 5) (s + 1)



Examples

Example


y ′′ − 2y ′ + 5y = −8e−t

y (0) = 2y ′ (0) = 12


2 We have Y (s) =2s2 + 10s

(s + 1) (s2 − 2s + 5)3 Earlier, we found that

L−1{

2s2 + 10s(s2 − 2s + 5) (s + 1)



Examples

Example


y ′′ + 4y ′ − 5y = tet

y (0) = 1y ′ (0) = 0


2 We have Y (s) =s3 + 2s2 − 7s + 5(s − 1)3 (s + 5)

=

181216 (s − 1) −

1

36 (s − 1)2+

1

6 (s − 1)3+

35216 (s + 5)

3 Hence,

y (t) = L−1{

181216 (s − 1) −

1

36 (s − 1)2+

1

6 (s − 1)3+

35216 (s + 5)

}4 Finally, y (t) =

181216

et − 136tet +

112t2et +

35216

e−5t


Examples

Example


y ′′ + 4y ′ − 5y = tet

y (0) = 1y ′ (0) = 0


2 We have Y (s) =s3 + 2s2 − 7s + 5(s − 1)3 (s + 5)

=

181216 (s − 1) −

1

36 (s − 1)2+

1

6 (s − 1)3+

35216 (s + 5)

3 Hence,

y (t) = L−1{

181216 (s − 1) −

1

36 (s − 1)2+

1

6 (s − 1)3+

35216 (s + 5)

}

4 Finally, y (t) =181216

et − 136tet +

112t2et +

35216

e−5t


Examples

Example


y ′′ + 4y ′ − 5y = tet

y (0) = 1y ′ (0) = 0


2 We have Y (s) =s3 + 2s2 − 7s + 5(s − 1)3 (s + 5)

=

181216 (s − 1) −

1

36 (s − 1)2+

1

6 (s − 1)3+

35216 (s + 5)

3 Hence,

y (t) = L−1{

181216 (s − 1) −

1

36 (s − 1)2+

1

6 (s − 1)3+

35216 (s + 5)

}4 Finally, y (t) =

181216

et − 136tet +

112t2et +

35216

e−5t


Examples

Example


w ′′ (t)− 2w ′ (t) + 5w (t) = −8eπ−t

w (π) = 2w ′ (π) = 12

1 The initial condition is not given at t = 0, what can we do?

2 Maybe a change of variable? Which one?3 Don’t forget to write the solution in terms of the original variable.4 Final answer: w (t) = 3et−π cos 2t + 4et−π sin 2t − e−t+π


Examples

Example


w ′′ (t)− 2w ′ (t) + 5w (t) = −8eπ−t

w (π) = 2w ′ (π) = 12

1 The initial condition is not given at t = 0, what can we do?2 Maybe a change of variable? Which one?

3 Don’t forget to write the solution in terms of the original variable.4 Final answer: w (t) = 3et−π cos 2t + 4et−π sin 2t − e−t+π


Examples

Example


w ′′ (t)− 2w ′ (t) + 5w (t) = −8eπ−t

w (π) = 2w ′ (π) = 12

1 The initial condition is not given at t = 0, what can we do?2 Maybe a change of variable? Which one?3 Don’t forget to write the solution in terms of the original variable.

4 Final answer: w (t) = 3et−π cos 2t + 4et−π sin 2t − e−t+π


Examples

Example


w ′′ (t)− 2w ′ (t) + 5w (t) = −8eπ−t

w (π) = 2w ′ (π) = 12

1 The initial condition is not given at t = 0, what can we do?2 Maybe a change of variable? Which one?3 Don’t forget to write the solution in terms of the original variable.4 Final answer: w (t) = 3et−π cos 2t + 4et−π sin 2t − e−t+π


Examples

Example


y ′′ + 2ty ′ − 4y = 1

y (0) = 0y ′ (0) = 0


2 Applying the Laplace transform gives Y ′ +(3s− s2

)Y =

−12s2

3 This is a linear equation in Y with solution Y =1s3+Ce

s2

4

s34 From a homework problem, what is lim

s→∞Y (s), what does it mean for

C?

5 Final answer: y (t) =12t2


Examples

Example


y ′′ + 2ty ′ − 4y = 1

y (0) = 0y ′ (0) = 0



)Y =

−12s2


s2

4

s3

4 From a homework problem, what is lims→∞

Y (s), what does it mean for

C?



Examples

Example


y ′′ + 2ty ′ − 4y = 1

y (0) = 0y ′ (0) = 0



)Y =

−12s2


s2

4

s34 From a homework problem, what is lim

s→∞Y (s), what does it mean for

C?



Exercises

Do the problems at the end of section 4.4 in my notes on solving initialvalue problems with the Laplace transform.


http://ksuweb.kennesaw.edu/~plaval/math2306/lap_ivp.pdf

http://ksuweb.kennesaw.edu/~plaval/math2306/lap_ivp.pdf

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