Question

If y = f(x) is solution of \(\frac{{{d^2}y}}{{d{x^2}}} = 0\) with the boundary conditions y = 5 at x = 0, and \(\frac{{dy}}{{dx}} = 2\) at x = 10, f(15) = _____
1. 15
2. 25
3. 35
4. 5

Answer 1

Correct Answer - Option 3 : 35

Concept:

For a second-order differential function, the characteristic equation will have two roots D1 and D2. The roots can have three possible forms i.e.

• Real, distinct roots, D1 ≠ D2
• Complex roots, D1, D2 = a ± ib
• Real and equal, D1 = D2 = D.

The solution for the second-order differential equation with equal roots of the characteristic equation is given by:

\(y = \left( {{C_1} + {C_2}x} \right){e^{{Dx}}}\)

\(\frac{{{d^2}y}}{{d{x^2}}} = 0 \Rightarrow {D^2}y = 0\)

D = 0 ⇒ D = 0, 0

y = (C1 + C2x) e0x ⇒ y = C1 + C2x ……(1)

Applying boundary conditions

At x = 0, y = 5

⇒ 5 = C1 + C2(0) ⇒ C1 = 5

And at x = 10, \(\frac{{dy}}{{dx}} = 2\)

\(\frac{{dy}}{{dx}} = C_2\)

Then form the equation (1), we have

y = 5 + 2x

Then, at x = 15, y = 5 + 2(15) = 35