\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.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 4: egin{align} y= sqrt{1+x^2} : y^{'}=frac{xy}{1+x^2}end{align}....
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