Laplace Transform for Solving a Second-Order ODE with Piecewise Functions

[manenbu]

Homework Statement

Using laplace transform, solve:

<br /> y&#039;&#039; - y = \frac{x^2-3x+2}{|x^2-3x+2|}

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

Homework Equations

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&#039;&#039; - y = 1 -2u_{1}(x) + 2u_{2}(x)
?


Will this be correct?

 

[gabbagabbahey]

Are u_1(x) and u_2(x) supposed to represent the Heaviside step functions

u_1(x)=\left\{\begin{array}{lr}0, &amp; x&lt;1 \\ 1, &amp; x\geq 1\end{array}\right. \;\;\;\;\;\;\;\;\; u_2(x)=\left\{\begin{array}{lr}0, &amp; x&lt;2 \\ 1, &amp; 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&#039;&#039;-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.