Hello,

I have the DE....

$\displaystyle y''(t) + \lambda^2 y(t) - x(t) = 0$

I know that

$\displaystyle y''(t) + \lambda^2 y(t) = 0$

gives $\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t$

The solution in the text gives

$\displaystyle y(t) = A \cos \lambda t + B \sin \lambda t + \frac{x(t)}{\lambda^2}$

but Im not sure why the solution has the $\displaystyle \frac{x(t)}{\lambda^2}$ part.