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?

  • $\begingroup$ You should give more details about what you tried. $\endgroup$ – Yves Daoust Sep 25 '16 at 13:25
  • 1
    $\begingroup$ If $a>1$, then the roots are complex. In this case there may be nontrivial solutions. And that's what the solution says. $\endgroup$ – Hans Engler Sep 25 '16 at 13:29

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


So there are solutions when $\sqrt{a-1}$ is an integer.

For the double root ($a=1$),

$$y=e^{-t}(px+q)$$ and


  • $\begingroup$ Crossed with @RobertZ. $\endgroup$ – Yves Daoust Sep 25 '16 at 13:34

Hint. Consider the case when $1-a>0$, $1-a<0$, $1-a=0$. Note that when $1-a<0$, then the general solution is $$y(x)=Ae^{-x}\cos(\sqrt{a-1}x)+Be^{-x}\sin(\sqrt{a-1}x).$$ The condition $y(0)=0$ implies that $A=0$ and $y(\pi)=0$ implies $$Be^{-\pi}\sin(\sqrt{a-1}\pi)=0$$ The solution is non trivial if $B\not=0$, therefore $\sqrt{a-1}$ should be an integer.


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