# Laplace transform for an ODE

manenbu

## Homework Statement

Using laplace transform, solve:

$$y'' - y = \frac{x^2-3x+2}{|x^2-3x+2|}$$

y(0)=y'(0)=0

## The Attempt at a Solution

Just to know if I'm on the right path -
Since the quadratic has 2 roots, at 1 and at 2, the entire thing can be equal to either 1 or -1, depends on x.
So can I write it as:
$$y'' - y = 1 -2u_{1}(x) + 2u_{2}(x)$$
?

Will this be correct?

## Answers and Replies

Homework Helper
Gold Member
Are $u_1(x)$ and $u_2(x)$ supposed to represent the Heaviside step functions

$$u_1(x)=\left\{\begin{array}{lr}0, & x<1 \\ 1, & x\geq 1\end{array}\right. \;\;\;\;\;\;\;\;\; u_2(x)=\left\{\begin{array}{lr}0, & x<2 \\ 1, & x\geq 2\end{array}\right.$$

???

If so, you need to be careful to exclude the two roots from the Domain of your final solution since $y''-y$ is indeterminant there. Other than that, it looks good to me.

manenbu
Yes, this is exactly what I meant.
Thank you - just making sure I wasn't doing something stupid.