# Solving IVPs with Laplace Transform

You've probably asked yourself why the Laplace transform is in the Differential Equations section. The answer is simple: because we can solve initial-value problems with the help of the Laplace transform.

Let's see how it is done.

Example 1. Solve y''+4y=t, y(0)=0, y'(0)=0.

The first step is always to take the Laplace transform of both sides.

L(y''+4y)=L(t),

Or

L(y'')+4L(y)=1/s^2.

Now, since L(y'')=s^2Y(s)-sy(0)-y'(0) (see table of Laplace transforms), we obtain that

s^2Y(s)-sy(0)-y'(0)+4Y(s)=1/s^2, or, using the initial conditions,

s^2Y(s)-s*0-0+4Y(s)=1/s^2, which yields Y(s)=1/(s^2(s^2+4)).

So, to find y(t), we have to find the inverse Laplace transform.

The partial fraction decomposition is A/s+B/s^2+(Cs+D)/(s^2+4)=(As(s^2+4)+B(s^2+4)+(Cs+D)s^2)/(s^2(s^2+4))=((A+C)s^3+(B+D)s^2+4As+4B)/(s^2(s^2+4))=1/(s^2(s^2+4))

So, the following system is obtained:

{(A+C=0),(B+D=0), (4A=0),(4B=1):}, which has the solution A=0, B=1/4, C=0, D=-1/4.

Thus,

y(t)=L^-1(Y(s))=L^(-1)(1/(s^2(s^2+4)))=L^-1(1/4 1/(s^2)-1/4 1/(s^2+4))=1/4 L^-1(1/s^2)-1/4 L^-1(1/(s^2+4))=1/4 t -1/8 L^-1(2/(s^2+4))=1/4t-1/8 sin(2t)

That's all. For linear differential equations, it is always the case that we take the Laplace transform, algebraically find Y(s), and take the inverse transform to obtain the solution.

Also, it is easier to solve IVPs that involve a step function or a Dirac function with Laplace transforms.

Example 2. Solve y''+y'=u_1(t), y(0)=0, y'(0)=0.

Take the Laplace transform

L(y'')+L(y')=L(u_1(t)),

Or

(s^2Y(s)-sy(0)-y'(0))+(sY(s)-y(0))=(e^-s)/s.

Applying the initial conditions gives:

s^2Y(s)+sY(s)=e^(-s)/s or Y(s)=e^(-s) 1/(s^2(s+1)).

The partical fraction decomposition for 1/(s^2(s+1)) is 1/s^2-1/s+1/(s+1).

So,

y(t)=L^-1(e^(-s) 1/s^2)-L^-1(e^(-s) 1/s)+L^-1(e^(-s) 1/(s+1))=u_1(t)(t-1)-u_1(t)1+u_1(t)e^(-(t-1))=u_1(t)(t-2+e^(1-t))

Of course, we could rewrite the Heaviside function by definition and solve the differential equation on each interval separately, but this would require a greater amount of work.

Now, let's consider one more example.

Example 3. Calculate y''+y'=t, y(2)=0, y'(2)=0.

Notice that the initial conditions are not at 0. In order to use the Laplace transform, we need the initial conditions to be at 0. For this, use the change of variable: let a=t-2; then:

y''(a+2)+y'(a+2)=a+2, y(0)=0, y'(0)=0

To simplify the expression, let y(a+2)=u(a); then, y'(a+2)=u'(a), and y''(a+2)=u''(a); so, the equation can be rewritten as:

u''+u'=a+2, u(0)=0, u'(0)=0.

Now, we can use the Laplace transform:

L(u'')+L(u')=L(a)+L(2)

s^2U(s)-su(0)-u'(0)+sU(s)-u(0)=1/s^2+2/s

Applying the initial conditions gives:

U(s)=(1/s^2+2/s)/(s^2+s)=(2s+1)/(s^3(s+1))

The partial fraction decomposition is (verify!): 1/s^3+1/s^2+1/(s+1)-1/s.

So, u(a)=L^-1(1/s^3+1/s^2+1/(s+1)-1/s)=L^-1(1/2 2/s^3+1/s^2+1/(s+1)-1/s)=1/2a^2+a+e^(-a)-1.

However, we need y(t), not u(a). Recall that y(t)=y(a+2)=u(a)=u(t-2); so,

y(t)=1/2 (t-2)^2+t-2+e^(-(t-2))-1=1/2 t^2-t-1+e^(2-t).

The Laplace transform can be used to solve linear equations with non-constant coefficients. In general, it is very hard to solve them, and the Laplace transform can rarely help, however such cases do exist.