Answer :
[tex]y''+\lambda y=0[/tex]
has the corresponding characteristic equation (CE)
[tex]r^2+\lambda=0[/tex]
a. If [tex]\lambda=0[/tex], then the CE has one root, [tex]r=0[/tex], and so the general solution to the ODE is
[tex]y(t)=C_1+C_2t[/tex]
Given that [tex]y'(0)=y'\left(\frac\pi6\right)=0[/tex], and
[tex]y'(t)=C_2[/tex]
it follows that [tex]C_2=0[/tex], and so
[tex]\boxed{y(t)=C_1}[/tex]
b. If [tex]\lambda>0[/tex], then the CE has two complex roots, [tex]r=\pm i\sqrt\lambda[/tex], and the general solution is
[tex]y(t)=C_1\cos(\lambda t)+C_2\sin(\lambda t)[/tex]
[tex]\implies y'(t)=-\lambda C_1\sin(\lambda t)+\lambda C_2\cos(\lambda t)[/tex]
With the given boundary values, we have
[tex]y'(0)=0\implies\lambda C_2=0\implies C_2=0[/tex]
[tex]y'\left(\dfrac\pi6\right)=0\implies-\lambda C_1\sin\left(\dfrac{\lambda\pi}6\right)=0[/tex]
[tex]\implies\sin\left(\dfrac{\lambda\pi}6\right)=0[/tex]
[tex]\implies\dfrac{\lambda\pi}6=n\pi[/tex]
[tex]\implies\lambda=6n[/tex]
where [tex]n\in\Bbb Z[/tex].
- If [tex]\lambda[/tex] is a (positive) multiple of 6, we have
[tex]y'\left(\dfrac\pi6\right)=0\implies-6nC_1\sin\left(\dfrac{6n\pi}6\right)=0\implies C_1=0[/tex]
and the solution would be
[tex]\boxed{y(t)=0}[/tex]
- Otherwise, if [tex]\lambda[/tex] is not a multiple of 6, we have
[tex]y'\left(\dfrac\pi6\right)=0\implies-\lambda C_1\sin\left(\dfrac{\lambda\pi}6\right)=0\implies C_1=0[/tex]
so that we still get
[tex]\boxed{y(t)=0}[/tex]
c. If [tex]\lambda<0[/tex], then the CE has two real roots, [tex]r=\pm\sqrt\lambda[/tex], so that the general solution is
[tex]y(t)=C_1e^{\sqrt\lambda\,t}+C_2e^{-\sqrt\lambda\,t}[/tex]
[tex]\implies y'(t)=C_1\sqrt\lambda\,e^{\sqrt\lambda\,t}-C_2\sqrt\lambda\,e^{-\sqrt\lambda\,t}[/tex]
From the boundary conditions we get
[tex]y'(0)=0\implies C_1\sqrt\lambda-C_2\sqrt\lambda=0\implies C_1=C_2[/tex]
[tex]y'\left(\dfrac\pi6\right)=0\implies C_1\sqrt\lambda\,e^{(\pi\sqrt\lambda)/6}-C_2\sqrt\lambda\,e^{-(\pi\sqrt\lambda)/6}=0\implies C_1e^{(\pi\sqrt\lambda)/3}=C_2[/tex]
from which it follows that [tex]C_1=C_2=0[/tex], so again the solution is
[tex]\boxed{y(t)=0}[/tex]
d. We only get eigenvalues in the case when [tex]\lambda>0[/tex], as in part (b):
[tex]\boxed{\lambda=6n,\,n\in\{1,2,3,\ldots\}}[/tex]
for which we get the corresponding eigenfunctions
[tex]\boxed{y(t)=\cos(6nt),\,n\in\{1,2,3,\ldots\}}[/tex]