Finding Eigen Values of ODE

Question

I am asked to find non-trivial eigen values of the BVP $$y''(x)+\lambda y(x) =0, \quad y \left(\frac{\pi}{2} \right) = 0, \quad y(0)= -3y'(0).$$ However, after few routine calculations, I obtained $\tan ^2 (\sqrt{\lambda} \pi/2) = 9\lambda$. It is also given that the eigen values may be given by $\lambda_n = 4n^2 \pi ^2$. I am unable to show the given solutions satisfy the equation $\tan ^2 (\sqrt{\lambda} \pi/2) = 9\lambda$.

Answer 1

For ease of notation, let $\lambda = \omega^2$, then the general solution is

$$ y(x) = A\cos(\omega x) + B\sin(\omega x) $$

The first B.C. gives

$$ y(0) = -3y'(0) \implies A = -3\omega B $$ $$ \implies Y(x) = B\big[{-3}\omega\cos(\omega x) + \sin(\omega x)\big] $$

The second B.C. gives

$$ y\left(\frac{\pi}{2}\right) = 0 \implies {-3}\omega\cos\left(\frac{\omega\pi}{2}\right) + \sin\left(\frac{\omega\pi}{2}\right) \\ \implies 3\omega = \tan\left(\frac{\omega\pi}{2}\right) $$

These eigenvalues do not have a closed form and will need to be solved numerically. However, as $\omega \to \infty$, you have the asymptotic approximation

$$ \cos\left(\frac{\omega\pi}{2}\right) - \frac{1}{3\omega}\sin\left(\frac{\omega\pi}{2}\right) \approx \cos\left(\frac{\omega\pi}{2}\right) = 0 $$

The solutions will we very close to the zeroes of the cosine function, hence $$ \omega_{n\to\infty} \approx 2n+1 $$

For very large values of $\omega$