The solution of the differential equation d 2 y d x 2 = sin 3 x + e x + x 2 when y 1 0 = 1 and y 0 = 0 , is

Question

TestHub · Accepted Answer

Integrating the given differential equation, we have

$\frac{d y}{d x} = \frac{- \cos 3 x}{3} + e^{x} + \frac{x^{3}}{3} + C_{1}$

But $y_{1} (0) = 1$

So, $1 = (- \frac{1}{3}) + 1 + C_{1}$

$\Rightarrow C_{1} = \frac{1}{3}$

$∴ \frac{d y}{d x} = \frac{- \cos 3 x}{3} + e^{x} + \frac{x^{3}}{3} + \frac{1}{3}$

Again integrating, we get

$y = - \frac{\sin 3 x}{9} + e^{x} + \frac{x^{4}}{12} + \frac{1}{3} x + C_{2}$

But $y (0) = 0,$ so $0 = 0 + 1 + C_{2}$

$\Rightarrow C_{2} = - 1$

Thus, $y = - \frac{\sin 3 x}{9} + e^{x} + \frac{x^{4}}{12} + \frac{1}{3} x - 1$

Hence, option (a) correct.

Mathematics - Differential Equation Question with Solution | TestHub