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

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Department
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