# Show that

Question:

$y=x \sin x \quad: x y^{\prime}=y+x \sqrt{x^{2}-y^{2}}(x \neq 0$ and $x>y$ or $x<-y)$

Solution:

$y=x \sin x$

Differentiating both sides of this equation with respect to x, we get:

$y^{\prime}=\frac{d}{d x}(x \sin x)$

$\Rightarrow y^{\prime}=\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)$

$\Rightarrow y^{\prime}=\sin x+x \cos x$

Substituting the value ofin the given differential equation, we get:

L.H.S. $=x y^{\prime}=x(\sin x+x \cos x)$

$=x \sin x+x^{2} \cos x$

$=y+x^{2} \cdot \sqrt{1-\sin ^{2}} x$

$=y+x^{2} \sqrt{1-\left(\frac{y}{x}\right)^{2}}$

$=y+x \sqrt{y^{2}-x^{2}}$

$=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.