Lesson 28 - Solving Linear Systems with Laplace Transforms

We can solve systems of linear equations containing derivatives using Laplace transforms.

Consider the linear system:              {x′−x−y=1,x(0)=0(1)−x+y′−y=0,y(0)=−52(2)

Notice that each equation contains a mix of x and y. This process is going to work similarly to solving linear systems of equations from algebra and precalculus classes. First, we are going to take the Laplace transform of each linear equation. We must recall the Laplace transforms of f(t).

L[f(t)]=F(s)
L[f′(t)]=sF(s)−f(0)

For the purpose of organization, let's label each equation:

Equation 1: x′−x−y=1, x(0)=0
Equation 2:−x+y′−y=0, y(0)=−52

Laplace transform for equation (1):

L[x′−x−y=1]→sX(s)−x(0)−X(s)−Y(s)=1s

Laplace transform for equation (2):

L[−x+y′−y=0]→−X(s)+sY(s)−y(0)−Y(s)=0

Initial conditions for equation (1):

sX(s)−0−X(s)−Y(s)=1ssX(s)−X(s)−Y(s)=1sX(s)(s−1)−Y(s)=1s

Initial conditions for equation (2):

−X(s)+sY(s)−(−52)−Y(s)=0−X(s)+sY(s)+52−Y(s)=0−X(s)+sY(s)−Y(s)+52=0−X(s)+Y(s)(s−1)=−52

Now let's look at our transformed system

{X(s)(s−1)−Y(s)=1s(1)−X(s)+Y(s)(s−1)=−52(2)

The goal now it to solve for one of Y(s) or X(s). This can be done in multiple ways. We can either multiply equation 1 by (s−1) or multiply equation 2 by (s−1). Let's multiply equation 2 to get rid of the X(s) terms.

{X(s)(s−1)−Y(s)=1s(1)(s−1)(−X(s)+Y(s)(s−1))=(s−1)(−52)(2){X(s)(s−1)−Y(s)=1s(1)−X(s)(s−1)+Y(s)(s−1)2=−5(s−1)2(2)

Now we can combine the linear systems by adding them. Notice the X(s) terms cancel out.

Y(s)(s−1)2−Y(s)=−5(s−1)2+1s

Now, we can solve this as we would any other Laplace transform differential equation.

Simplifying the right side:

Y(s)((s−1)2−1)=−5(s−1)2+1sY(s)((s−1)2−1)=−5s−52+1sY(s)((s−1)2−1)=(ss)−5s−52+1s(22)Y(s)((s−1)2−1)=−5s2−5s+22s

Expanding the left side:

Y(s)(s2−2s+1−1)=−5s2−5s+22sY(s)(s2−2s)=−5s2−5s+22s

Solving for Y(s)

Y(s)=−5s2−5s+22s(s2−2s)Y(s)=−5s2−5s+22s2(s−2)

I will leave you to solve for the partial fractions here.

You should end up with

12[−5s2−5s+2s(s2−2s)]=As+Bs2+Cs−2

with A=−1, B=−3, and C=−2. (Note: Pulling out the 12 makes it easier to see the decomposition structure.)

Now take the inverse Laplace transform. Don't forget about the 12 from earlier.

y(t)=12[L−1[−1s]−3L−1[1s2]−2L−1[1s−2]]y(t)=−t2−32−e2t

Now, we have y(t). However, we're not quite done as we need to solve for x(t) as well. We can utilize y(t) to come up with that solution using one of our original equations.

If we choose equation (1), we only know y(t). If we use equation (2), we know y(t) and we can find y′(t) by taking the derivative of y(t). For this reason, let's use equation (2).

y(t)=−t2−32−e2ty′(t)=−12−2e2t

Now, let's utilize equation 2 and plug in the values.

−x−12−2e2t−[−t2−32−e2t]=0

Now, we solve for x(t).

−x(t)−12−2e2t+t2+32+e2t=0x(t)=1+t2−e2t

Now here's our final solution:

y(t)=−t2−32−e2tx(t)=t2+1−e2t

Try for Yourself

Solve the linear system:     {x′+y=0,x(0)=744x+y′=3,y(0)=4
Solve the linear system:     {x′+y=1−u(t−2),x(0)=0x+y′=0,y(0)=0
Solve the linear system:     {x′−3x+2y=sin⁡t,x(0)=04x−y′−y=cos⁡t,y(0)=0

Next Lesson: Lesson 29 - Quick Lesson on Matrices

Table of Contents: Table of Contents