Question

Consider the following statements
I. The function $y=x\sin 3x$ is a solution of the differential equation $\frac{d^{2}y}{d{x}^2}+9y-6\cos 3x=0$.
II. The function $x^2=2y^2\log y$ is a solution of the differential equation $\left(x^2+y^2\right)\frac{dy}{dx}-xy=0$.
Choose the correct option.

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (C)

I. Given, $y=x\sin 3x.....$(i)
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=x\frac{d}{dx}(\sin 3x)+\sin 3x\frac{d}{dx}(x)$
(using product rule of differentiation)
$\frac{dy}{dx}=x\cos 3x\times 3+\sin 3x$
Again, differentiating both sides w.r.t. $x$, we get
$\frac{d^{2}y}{d{x}^2}=3\left[x\frac{d}{dx}\cos 3x+\cos 3x\frac{d}{dx}(x)\right]+\frac{d}{dx}(\sin 3x)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=3[x(-\sin 3x)\cdot (3)+\cos 3x]+(\cos 3x)\cdot (3)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-9x\sin 3x+3\cos 3x+3\cos 3x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-9x\sin 3x+6\cos 3x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-9y+6\cos 3x\text{ [using Eq. (i)] }$
$\Rightarrow \frac{d^{2}y}{d{x}^2}+9y-6\cos 3x=0$
Hence, the given function is a solution of the corresponding differential equation.
II. Given, $x^2=2y^2\log y....$(i)
On differentiating both sides w.r.t. $x$, we get
$2x=2\left[y^2\times \frac{1}{y}\frac{dy}{dx}+\log y\times 2y\frac{dy}{dx}\right]$
$\Rightarrow 2x=2\{y+2y\log y\}\frac{dy}{dx}$
$\Rightarrow x=(y+2y\log y)\frac{dy}{dx}$
On multiplying both sides by $y$, we get
$xy=\left(y^2+2y^2\log y\right)\frac{dy}{dx}$
$\Rightarrow xy=\left(y^2+x^2\right)\frac{dy}{dx}$[using Eq. (i)]
$\Rightarrow \left(x^2+y^2\right)\frac{dy}{dx}-xy=0$
Hence, the given function is a solution of the corresponding differential equation.