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.