u''(x) +(-2 -x^2 )u = 0 ,\quad 0
I'm not sure where to even start. What method?
u''(x) +(-2 -x^2 )u = 0 ,\quad 0
I'm not sure where to even start. What method?
$u\equiv0$ is a solution. Are there any others? The answer is no. To see this, multiply the equation by $u$ and integrate between $0$ and $1$, to get \int_0^1u''(x)\,u(x)\,dx=\int_0^1(2+x)^2\,(u(x))^2\,dx. The right hand side is non-negative. Integrating by parts and using the boundary conditions, we see that the left hand side is non-positive. It follows that $ \int_0^1(2+x)^2\,(u(x))^2\,dx=0 $ and $u\equiv0$.
This argument works for any equation u''(x)=V(x)\,u(x) with $V(x)\ge0$ and boundary conditions $u(0)=u(1)=0$, u(0)=u'(1)=0, u'(0)=u(1)=0, or u'(0)=u'(1)=0.