Question

The second smallest eigenvalue of the eigenvalue problem \[ \frac{d^2 y}{dx^2} + (\lambda - 3)\,y = 0,\qquad y(0)=y(\pi)=0, \] is

• A. 4
• B. 3
• C. 7
• D. 9

Hint:

For $y''+(\lambda-3)y=0$ with $y(0)=y(\pi)=0$, set $\mu^2=\lambda-3$; the Dirichlet eigenvalues are $\,\mu=n\,$ giving $\,\lambda_n=n^2+3\,$. Count up to the requested order.

Answer 1

The Correct Option is C

Step 1: Reduce to the standard sine–cosine form.
Let $\mu^2=\lambda-3$. Then the ODE becomes \[ y'' + \mu^2 y = 0. \] General solution: \[ y(x)=A\cos(\mu x)+B\sin(\mu x). \]
Step 2: Apply boundary conditions.
$y(0)=0 \Rightarrow A=0$; hence $y(x)=B\sin(\mu x)$.
$y(\pi)=0 \Rightarrow B\sin(\mu \pi)=0$. For a nontrivial solution ($B\neq 0$), we need \[ \sin(\mu \pi)=0 \Rightarrow \mu = n, n=1,2,3,\ldots \]
Step 3: Recover $\lambda_n$.
Since $\mu^2=\lambda-3$, \[ \lambda_n = \mu^2 + 3 = n^2 + 3,\qquad n=1,2,3,\ldots \] Thus the eigenvalues in increasing order are \[ \lambda_1=1^2+3=4, \lambda_2=2^2+3=7, \lambda_3=3^2+3=12,\ldots \] The {second smallest} is $\lambda_2=7$. Final Answer: \[ \boxed{7} \]