Originally Posted by

**Ulysses** Hi there. I have to solve the differential equation using the laplace transform.

$\displaystyle ty''+2y'+ty=0,y(0)=1,y(\pi)=0$

So, applying the Laplace transform on both sides of the equality, and using some properties of the Laplace transform I get:

$\displaystyle -\frac{d}{ds}(s^2y(s)-s-y'(0))+2sy(s)-2y(0)-\frac{dy}{ds}=0$

$\displaystyle -s^2\frac{dy}{ds}-\frac{dy}{ds}-1=0\rightarrow -\frac{ds}{s^2+1}=dy$

Then: $\displaystyle y(s)=-\arctan(s)+C$

Then I have to apply the inverse transform to get the function I'm looking for, but the thing is I don't know the inverse Laplace transform for the arctangent of s. I think I have probably done something wrong, but I couldn't find the mistake.