Question

The solution of the differential equation $\frac{d^{2}y}{d{x}^2}=\sin 3x+e^x$ $+x^2$ when $y^{\prime }(0)=1$ and $y(0)=0$ is

• A. $\frac{-\sin 3x}{9}+e^x+\frac{x^4}{12}+\frac{1}{3}x-1$
• B. $\frac{-\sin 3x}{9}+e^x+\frac{x^4}{12}+\frac{1}{3}x$
• C. $\frac{-\cos 3x}{3}+e^x+\frac{x^4}{12}+\frac{1}{3}x+1$
• D. None of these

Answer 1

Answer: (A)

Integrating the given differential equation, we have
$\frac{dy}{dx}=\frac{-\cos 3x}{3}+e^x+\frac{x^3}{3}+C_1$ but $y^{\prime }(0)=1$
so $1=\left(-\frac{1}{3}\right)+1+C_1$
$\Rightarrow C_1=1/3$.
Again integrating, we get
$y=\frac{-\sin 3x}{9}+e^x+\frac{x^4}{12}+\frac{1}{3}x+C_2$
but $y(0)=0$ so $0=0+1+C_2$
$\Rightarrow C_2=-1$. Thus
$y=\frac{-\sin 3x}{9}+e^x+\frac{x^4}{12}+\frac{1}{3}x-1$