# Solving IVP's with Laplace Transform

You probably asked yourself why Laplace transform is in Differential Equations section. Answer is simple. Because we can solve initial-value problems with the help of Laplace transform.

Let's see how it is done.

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

First step always is to take 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) then we obtain that

s^2Y(s)-sy(0)-y'(0)+4Y(s)=1/s^2 or using 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 must find inverse Laplace transform.

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 solution A=0, B=1/4, C=0, D=-1/4

So,

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

Also, it is easier to solve IVP's that involve step function or dirac function with Laplace transforms.

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

Take 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 initial conditions gives

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

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 Heaviside function by definition and solve differential equation on each interval separately, but this requires more amount of work.

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

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

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

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

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

Now we can use 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 initial conditions gives

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

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)

Laplace transform can be used to solve linear equations with non-constant coefficients, but in general it is very hard to solve them and Laplace transform can rearely help, however such cases exist.