Non trivial solutions to $ y'' + 2y' + ay = 0$ For which values of $a$ does the following equation has non trivial solutions
$y'' + 2y' + ay = 0 , \space y(0) = y(\pi ) = 0$
The characteristic equation is:
$$x^2+2x+a = 0$$
and I have found the roots to be
$$x_1 = \sqrt{(1-a)}-1$$
$$x_2 = -\sqrt{(1-a)}-1$$
I have tried for $a =1, a<1, a>1$.
$a=1$, I got: $y(t) = C_1e^{-t}+C_2e^{-t}$ and $C_1=C_2=0$
$a>1$, gives complex roots and the solution is of the style:
$y(t) = C_1e^{-t}\cos b+C_2e^{-t}\sin b$ and $C_1=C_2=0$
$a<1$, gives distinct roots. 
I don't know if I have calculated the complex roots right.
I have found that all solutions give the trivial. Is that correct?
The solution says that $a=1+n^2$
Where does that come from?

From the roots, when they are real and distinct,
$$y=pe^{x_1x}+qe^{x_2x}$$ and
$$0=p+q=pe^{x_1\pi}+qe^{x_2\pi},\\0=p(e^{x_1x}-e^{x_2x})=pe^{x_1x}(1-e^{(x_2-x_1)x})$$ you indeed get the trivial solution.
But when they are complex ($a>1$),
$$y=e^{-x}(p\cos\sqrt{a-1}x+q\sin\sqrt{a-1}x)=$$ and
$$0=p=e^{-\pi}(p\cos\sqrt{a-1}\pi+q\sin\sqrt{a-1}\pi).$$
So there are solutions when $\sqrt{a-1}$ is an integer.

For the double root ($a=1$),
$$y=e^{-t}(px+q)$$ and
$$0=q=e^{-\pi}(p\pi+q).$$
