# Laplace transform for solving ODE with variable coefficients

1. Dec 4, 2006

### angelas

Can we use laplace transform to solve an ODE with variable coefficients?

Like this one:

4x y" + 2 y' + y = exp (-x)

2. Dec 4, 2006

### HallsofIvy

Staff Emeritus
Theoretically, yes, if you can find the Laplace transform of f(x)y"(x)!

The Laplace transform of xy"(x) is, by definition,
$$\int_0^\infty e^{-st}ty"(t)dt$$
Integrate by parts with $u= e^{-st}t$ so that $du= (e^{-st}- st e^{-st})dt$, dv= y"(t)dt so that v= y'(t). Then
$$-\int_0^\infty (e^{-st}- ste^{-st})y'(t)dt$$
Do that by integration by parts with $u= (e^{-st}- ste^{-st}$ so that $du= -2e^{-st}+ s^2te^{-st}$, dv= y'(t)dt so that v= y. Then
[tex]2y(0)+ \int_0^\infty (2e^{-st}- s^2te^{-st})y(t)dt[/itex]
Write that last as a Laplace transform of y and reduce to an algebraic equation as usual.

Unfortunately, that is not always easy to do with general variable coefficients. The Laplace transform is basically a method for very mechanically solving linear equations with constant coefficients.

3. Dec 4, 2006

### angelas

Last edited: Dec 4, 2006