Class 12 Mathematics - Chapter Differential Equations NCERT Solutions | \begin{align} y= \sqrt{1+x^2} :&nbs
\begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align}
\begin{align} y= \sqrt{1+x^2}\end{align}
Differentiating both sides of the equation with respect to x, we get:
\begin{align} y^{'}=\frac{d}{dx}\left(\sqrt{1+x^2} \right)\end{align}
\begin{align} y^{'}=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}\left(1+x^2\right)\end{align}
\begin{align} y^{'}=\frac{2x}{2\sqrt{1+x^2}}\end{align}
\begin{align} y^{'}=\frac{x}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{\sqrt{1+x^2}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{y}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{xy}{1+x^2}\end{align}
∴ L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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