Correct option is D
Step 1: Rearrange the equation
The equation can be written as:
y'' + (λ – 3) y = 0.
Let k² = λ – 3.
Step 2: Solve the differential equation
There are three cases:
If k² = 0 → λ = 3, then equation is y'' = 0.
General solution: y(x) = ax + b.
Apply y(0) = 0 → b = 0.
Apply y(π) = 0 → aπ = 0 → a = 0.
So only trivial solution. Not acceptable.
If k² < 0 → λ < 3, the solutions are hyperbolic (cosh, sinh).
With boundary conditions y(0) = 0, y(π) = 0, this forces trivial solution only. Not acceptable.
If k² > 0 → λ > 3, the solutions are sinusoidal:
y(x) = A sin(kx) + B cos(kx).
From y(0) = 0 → B = 0.
Then y(x) = A sin(kx).
Apply y(π) = 0 → A sin(kπ) = 0.
For non-trivial solution, A ≠ 0, so sin(kπ) = 0 → k = n, where n = 1, 2, 3,...
Step 3: Find eigenvalues
So, k = n → λ = 3 + n², for n = 1, 2, 3,...
The smallest value occurs when n = 1 → λ = 3 + 1² = 4.
Final Answer:
The smallest λ is 4
