Show that
Question:

$y=\sqrt{1+x^{2}} \quad: y^{\prime}=\frac{x y}{1+x^{2}}$

Solution:

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

Differentiating both sides of the equation with respect to x, we get:

$y^{\prime}=\frac{d}{d x}\left(\sqrt{1+x^{2}}\right)$

$y^{\prime}=\frac{1}{2 \sqrt{1+x^{2}}} \cdot \frac{d}{d x}\left(1+x^{2}\right)$

$y^{\prime}=\frac{2 x}{2 \sqrt{1+x^{2}}}$

$y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \times \sqrt{1+x^{2}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \cdot y$

$\Rightarrow y^{\prime}=\frac{x y}{1+x^{2}}$

L.H.S. = R.H.S.

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

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.