Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y=x sin x : xy'=$$y+x\sqrt{x^2-y^2}(x\neq 0\,and x>y\,or x<-y)

Answer 1

y=x sin x
Differentiating both sides of this equation with respect to x, we get:
$y'=\frac{d}{dx}(x\sin x)$

$\Rightarrow y'=\sin x.\frac{d}{dx}(x)+x.\frac{d}{dx}(\sin x)$

$y'=\sin x+ \cos x$
Substituting the value of y' in the given differential equation, we get:
L.H.S.= $xy'=x(\sin x+ \cos x)$

=x sin x+x2 cos x

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

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

=$y+x\sqrt{y^2-x^2}$
=R.H.S.

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