SOLUTION OF DIFFERENTIAL EQUATIONS BY LAPLACE TRANSFORMS

9. SOLUTION OF DIFFERENTIAL EQUATIONS BY LAPLACE TRANSFORMS

We are going to discuss the solution of linear differential equations with constant coefficients (see Sections 5 and 6). Laplace transforms can reduce such an equation to an algebraic equation and so simplify solving it. Also, since Laplace transforms automatically use given values of initial conditions, we find immediately a desired particular solution without the extra step of determining constants to satisfy the initial conditions. Discontinuous forcing functions are messy to deal with by Section

6 methods; the Laplace transform method handles them easily. We are going to take Laplace transforms of the terms in differential equations; to do this we need to know the transforms of derivatives y ′ = dy/dt, y ′′ =d 2 y/dt 2 ,

etc. To find L(y ′ ), we use the definition (8.1) and integrate by parts, as follows

(9.1) L(y ′ )=

y ′ (t)e −pt dt = e −pt y(t) − (−p)

y(t)e −pt dt

= −y(0) + pL(y) = pY − y 0

where for simplicity we have written L(y) = Y and y(0) = y 0 . To find L(y ′′ ), we think of y ′′ as (y ′ ) ′ , and substitute y ′ for y in (9.1) to get

L(y ′′ ) = pL(y ′ )−y ′ (0).

Using (9.1) again to eliminate L(y ′ ), we finally have (9.2)

L(y ′′ )=p 2 L(y) − py(0) − y ′ (0) = p 2 Y − py 0 −y ′ 0 . Continuing this process, we obtain the transforms of the higher-order derivatives

(Problem 1 and L35). We are now ready to solve differential equations. We illustrate the method by some examples.

Example 1. Solve y ′′ + 4y ′ + 4y = t 2 e −2t with initial conditions y 0 = 0, y 0 ′ = 0. We take the Laplace transform of each term in the equation, using L35 and L6 in the table of Laplace transforms. We get

But the initial conditions are y 0 =y 0 ′ = 0. Thus we have

. Now we want y, which is the inverse Laplace transform of Y . We look in the table

for the inverse transform of 2/(p + 2) 5 . By L6, we get

This is much simpler than the general solution; we have obtained just the solution satisfying the given initial conditions.

Section 9 Solution of Differential Equations by Laplace Transforms 441 Example 2. Solve y ′′ + 4y = sin 2t, subject to the initial conditions y 0 = 10, y ′ 0 = 0.

Using the table, take the Laplace transform of each term of the equation to get

p 2 Y − py 0 −y ′ 0 + 4Y = L(sin 2t) =

. p 2 +4

Then we substitute the initial conditions and solve for Y as follows:

Finally, taking the inverse transform using L4 and L17, we have the desired solution: y = 10 cos 2t + 1 8 (sin 2t − 2t cos 2t) = 10 cos 2t + 1 8 sin 2t − 1 4 t cos 2t.

Example 3. Solve y ′′ + 4y ′ + 13y = 20e −t , y 0 = 1, y ′ 0 = 3.

We take the transform of each term and solve for Y as follows: p 2 Y − p − 3 + 4pY − 4 + 13Y = 20 ,

p+1

1 20 p 2 + 8p + 27

2 p . + 4p + 13 p+1 (p + 1)(p + 4p + 13) Since this Y is not in our table, we can either use a larger table, or use partial

Y= 2 +p+7 =

fractions to split Y into fractions which are in our table (which you can do by computer) or find the inverse transform by computer. We find:

2 p+2 Y=

−p + 1

2 p+1 − p + 4p + 13 p+1 (p + 2) +9 (p + 2) 2 +9 and by L2, L13, and L14,

y = 2e −t +e −2t sin 3t − e −2t cos 3t.

Sets of simultaneous differential equations can also be solved by using Laplace transforms. Here is an example.

Example 4. Solve the set of equations

y ′ − 2y + z = 0, z ′ − y − 2z = 0,

subject to the initial conditions y 0 = 1, z 0 = 0.

We shall call L(z) = Z and L(y) = Y as before. We take the Laplace transform of each of the equations to get

pY − y 0 − 2Y + Z = 0, pZ − z 0 − Y − 2Z = 0.

442 Ordinary Differential Equations Chapter 8

After substituting the initial conditions and collecting terms, we have

(p − 2)Y + Z = 1, Y − (p − 2)Z = 0.

We solve this set of algebraic equations simultaneously for Y and Z (by any of the methods usually used for a pair of simultaneous equations—elimination, determi- nants, etc.). For example, we may multiply the first equation by (p − 2) and add the second to get

We find y by looking up the inverse transform of Y using L14. We get

y=e 2t cos t.

Similarly, solving for Z and looking up the inverse transform, we find

. Alternatively, we could find z from the first differential equation by substituting the y solution:

z = 2y − y 2t ′ = 2e cos t + e sin t − 2e cos t = e sin t. Solving linear differential equations with constant coefficients is not the only use

of Laplace transforms. As you will see in Chapter 13, Section 10, we may solve some kinds of partial differential equations by Laplace transforms. Also a table of Laplace

transforms can be used to evaluate definite integrals of the type ∞ 0 e −pt f (t)dt. Example 5. By L15 with a = 3 and p = 2, we have

e −2t

(1 − cos 3t) dt =

26 Actually, there is more to the subject than this. Although we are discussing

in this chapter the use of Laplace transforms as a tool, they also can play a more theoretical role in applied problems. It is often possible to find desired information about a problem directly from the Laplace transform of the solution without ever finding the solution. Thus the use of Laplace transforms may lead to a better understanding of a problem or a simpler method of solution. (Compare the use of matrices, for example, or the use of Fourier transforms.)

PROBLEMS, SECTION 9

1. Continuing the method used in deriving (9.1) and (9.2), verify the Laplace trans- forms of higher-order derivatives of y given in the table (L35).

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

2. y ′ − y = 2e t ,

y 0 =3

Section 9 Solution of Differential Equations by Laplace Transforms 443

12. y ′′ −y=e − t − 2te − t ,

y 0 = 1, y ′ 0 =2

13. y ′′ + y = 5 sinh 2t,

y 0 = 0, y ′ 0 =2

14. y ′′

− 4y ′ = −4te 2t , y 0 = 0, y 0 =1

20. y ′′ − 8y ′ + 16y = 32t,

Solve the following sets of equations by the Laplace transform method. 27. y ′ +z ′ − 3z = 0

z ′ −y=1

z 0 =1

29. y ′ +z ′ − 2y = 1

y 0 =z 0 =1

z−y ′ =t 30. y ′ + 2z = 1

y 0 =0

2y − z ′ = 2t

z 0 =1

31. y ′′ +z ′′ −z ′ =0

y 0 = 0, y ′ 0 =1

y ′ +z t − 2z = 1 − e z 0 = 1, z

32. z ′ + 2y = 0

y 0 =z 0 =0

y ′ − 2z = 2

444 Ordinary Differential Equations Chapter 8

Evaluate each of the following definite integrals by using the Laplace transform table. Z ∞

. Hint: In (8.1), let p = 2, f (t) = sin 3t; use L3 with a = 3. Z ∞

40. −e −2et dt 0 t

41. e −2t sin(t

2 ) dt