Question
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
$\text{y} = \text{x} \ \text{sin} \ \text{x}\ : \ \text{xy}' = \text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2} $ $(\text{x} \neq 0 \ \text{and} \ \text{x} > \text{y} \ \text{or} \ \text{x} < – \text{y})$

Answer

Given: $y = x \sin x .....(i)$ To prove: $y$ given by eq. $(i)$ is a solution of differential equation $\text{xy}'=\text{y}+\text{x} \sqrt{\text{x}^2-\text{y}^2}\ ....(\text{ii})$
Proof: From eq. $(i),$
$\frac{\text{dy}}{\text{dx}}(=\text{y}') = \text{x} \frac{\text{d}}{\text{dx}}\text{sin}\ \text{x}+ \text{sin}\ \text{x} \frac{\text{d}}{\text{dx}} \text{x} = \text{x}\ \text{cos}\ \text{x}\ + {\text{sin}} \ \text{x}\text{L.H.S.}$ of eq. $(ii)$
$= xy' = x (x \cos x + \sin x) = x^2 \cos x + x \sin x \text{R.H.S.}$ of eq. $(ii)$
$= \text{y} + \text{x}\sqrt{\text{x}^2-\text{y}^2} = \text{x}\ \text{sin} \ \text{x}+\text{x} \sqrt{\text{x}^2-\text{x}^2\ \text{sin}^2\text{x}}$ [From eq. (i)]$= \text{x}\ \text{sin}\ \text{x}+ \text{x} \sqrt{\text{x}^2(1-\text{sin}^2\text{x})} = \text{x}\ \text{sin}\ \text{x}+\text{x}\sqrt{\text{x}^2\text{cos}^2\text{x}}$
$= x \sin x + x.x \cos x = x \sin x + x^2 \cos x$
$= x^2 \cos x + x \sin x$
$\therefore \text{L.H.S. = R.H.S}$
Hence, $y$ given by eq. $(i)$ is a solution of $\text{xy}'=\text{y}+\text{x}\sqrt{\text{x}^2-\text{y}^2}.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free