Class 12 Mathematics - Chapter Differential Equations NCERT Solutions | \begin{align} y = xsinx:xy{'}=y +x\s
\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}
y= x.sinx
Differentiating both sides of this equation with respect to x, we get:
\begin{align} y^{'} =\frac{d}{dx}\left(x.sinx\right)\end{align}
\begin{align}\Rightarrow y^{'} =sinx. \frac{d}{dx}\left(x\right)+ x. \frac{d}{dx}\left(sinx\right)\end{align}
\begin{align} \Rightarrow y^{'} =sinx + x.cosx\end{align}
Differentiating both sides of this equation with respect to x, we get:
L.H.S. =xy' = x(sinx + xcosx)
\begin{align} =x.sinx + x^2.cosx\end{align}
\begin{align} =y + x^2.\sqrt{1-sin^2x}\end{align}
\begin{align} =y + x^2.\sqrt{1-\left(\frac{y}{x}\right)^2}\end{align}
\begin{align} =y + x^2.\sqrt{\frac{x^2-y^2}{x^2}}\end{align}
\begin{align} =y + x.\sqrt{x^2-y^2}\end{align}
R.H.S.
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