## Solving Linear ODE Using Laplace Transforms

How can we use Laplace transforms to solve ode? The procedure is best
illustrated with an example. Consider the ode

This is a linear homogeneous ode and can be solved using standard methods.

Let Y(s)=L[y(t)](s). Instead of solving directly for y(t), we derive a
new equation for Y(s). Once we find Y(s), we inverse transform
to determine y(t).

The first step is to take the Laplace transform of both sides of the
original differential equation. We have

Obviously, the Laplace transform of the function 0 is 0. If we look at
the left-hand side, we have

Now use the formulas for the L[y'']and L[y']:

Here we have used the fact that y(0)=2. And,

Hence, we have

The Laplace-transformed differential equation is

This is a linear algebraic equation for Y(s)! We have converted a
differential equation into a algebraic equation! Solving for Y(s), we have

We can simplify this expression using the method of partial fractions:

Recall the inverse transforms:

Using linearity of the inverse transform, we have

Another Example

Consider the ode:

This problem has an inhomogeneous term. In the direct approach
one solves for the homogeneous solution and the particular solution
separately. For this problem the particular solution can be determined
using variation of parameters or the method of undetermined
coefficients. Using the Laplace transform technique we can solve for
the homogeneous and particular solutions at the same time.

Let Y(s) be the Laplace transform of y(t). Taking the Laplace transform
of the differential equation we have:

The Laplace transform of the LHS  L[y''+4y'+5y] is

The Laplace transform of the RHS is

Equating the LHS and RHS and using the fact that y(0)=1 y'(0)=2, we
obtain

Solving for Y(s), we obtain:

Using the method of partial fractions it can be shown that

Using the fact that the inverse of 1/(s-1) is e^t and that the inverse of
1/[(s+2)^2+1] is exp(2t)sin(t), it follows that