Question

The differential equation

\(\frac{{{d^2}y}}{{d{x^2}}} + 16y = 0\) For y(x) with the two boundary conditions \({\left. {\frac{{dy}}{{dx}}} \right|_{x = 0}} = 1\;and{\left. {\frac{{dy}}{{dx}}\;} \right|_{x = \frac{\pi }{2}}} = - 1\;\)
1. No solution
2. Exactly two solutions
3. Exactly one solution
4. Infinitely many solutions

Answer 1

Correct Answer - Option 1 : No solution

Concept:

Given equation is

\(\frac{{{d^2}y}}{{d{x^2}}} + 16y = 0\)

This is a homogeneous second order differential equation,

So (D² + 16)y = 0

D² = m²

⇒ m² + 16 = 0     ⇒ m = ± 4i = 0 ± 4i

Solution is given as in this case roots are complex, m = α ± i β

y = (C₁ cos βx + C₂ sin βx) e^αx

= (C₁ cos 4x + C₂ sin 4x) e^ox = C₁ cos 4x + C₂ sin 4x

Now y’ = -4C₁ sin 4x + 4C₂ cos 4x

Applying Boundary condition,

y’ (0) = 1  ⇒ -4C₁ sin (0) + 4C₂ cos(0) = 1

\(4{C_2} = 1 \Rightarrow {C_2} = \frac{1}{4}\)

Putting another boundary condition.

\(y'\left( {\frac{\pi }{2}} \right) = - 1\)

\( - 4{C_1}\sin \left( {\frac{{4\pi }}{2}} \right) + 4{{\rm{C}}_2}\cos \left( {\frac{{4\pi }}{2}} \right) = - 1 \Rightarrow {C_2} = \frac{{ - 1}}{4}\)

So this equation has no solution.