Question

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