Solving Linear ODE Using the Laplace Transform Method
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''](s) and L[y']:
Here we have used the fact that y(0)=1
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 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). Hence, the solution is
