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{\Large\bf Solving Linear ODE Using the Laplace Transform Method}

\vspace{.3in}

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

$$
y''-5y'+6y=0 \hspace{.2in}y(0)=2, y'(0)=2
$$

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

$$
L[y''-5y'+y](s)=L[0](s)
$$

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

$$
L[y''-5y'+6y](s)=L[y''](s)-5L[y'](s)+6L[y](s)
$$

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

$$
L[y']=sL[y]-y(0)=sY(s)-2
$$

Here we have used the fact that y(0)=1

$$
L[y'']=s^2L[y]-sy(0)-y'(0)=s^2Y(s)-2s-2
$$

Hence, we have

$$
L[y''](s)-5L[y'](s)+6L[y](s)=s^2Y(s)-2s-2-5sY(s)+10+6Y(s)
$$

The Laplace-transformed differential equation is

$$
(s^2-5s+6)Y(s)-2s-2+10=0
$$

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

$$
Y(s)=\frac{2s-8}{s^2-5s+6}
$$

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

$$
Y(s)=\frac{4}{s-2}+\frac{-2}{s-3}
$$

Recall the inverse transforms:

$$
L^{-1}[1/(s-2)](t)=e^{2t} \hspace{.2in} L^{-1}[1/(s-3)](t)=e^{3t}
$$

Using linearity of the inverse transform, we have

$$
y(t)=L^{-1}[\frac{4}{s-2}+\frac{-2}{s-3}](t)=4e^{2t}-2e^{3t}
$$

\vspace{.3in}

{\bf Another Example}

\vspace{.2in}

Consider the ode:

$$
y''+4y'+5y=e^t \hspace{.2in}y(0)=1 y'(0)=2
$$

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:

$$
L[y''+4y'+5y]=L[10e^t]
$$

The Laplace transform of the LHS is

$$
L[y''+4y'+5y]=L[y'']+4L[y']+5L[y]=s^2Y(s)-sy(0)-y'(0)+4sY(s)-4y(0)+5Y(s)
$$

The Laplace transform of the RHS is

$$
L[20e^t]=\frac{10}{s-1}
$$

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

$$
(s^2+4s+5)Y(s)-s-6=\frac{10}{s-1}
$$

Solving for Y(s), we obtain:

$$
Y(s)=\frac{s+6}{s^2+4s+5}+\frac{10}{(s-1)(s^2+4s+5)}
$$

Using the method of partial fractions it can be
shown that 

$$
Y(s)=\frac{1}{(s+2)^2+1}+\frac{1}{s-1}
$$

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).
Hence, the solution is

$$
y(t)=e^{2t}\sin t+e^t.
$$

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