Question

Find the particular solution of the following differential equation
$y\frac{dy}{dx}=\sqrt{1+x^2+y^2+x^2{y}^2}$
given that $y\left(0\right)=0$.

Answer 1

Lets take the given differential equation
$y\frac{dy}{dx}=\sqrt{1+x^2+y^2+x^2{y}^2}$

Rearrange the terms

$y\frac{dy}{dx}=\sqrt{1+x^2+y^2(1+x^2)}$
$y\frac{dy}{dx}=\sqrt{(1+y^2)(1+x^2)}$

$\frac{y}{\sqrt{(1+y^2)}}\frac{dy}{dx}=\sqrt{(1+x^2)}$

$\frac{y}{\sqrt{(1+y^2)}}dy=\sqrt{(1+x^2)}dx$

Integrating both the sides, we get

$\int \frac{y}{\sqrt{(1+y^2)}}dy=\int \sqrt{(1+x^2)}dx$

$\frac{1}{2}\int \frac{2y}{\sqrt{(1+y^2)}}dy=\int \sqrt{(1+x^2)}dx$

We know that$\int \frac{dt}{\sqrt{t}}$= 2$\sqrt{t}+C$
Therefore $\frac{1}{2}\int \frac{2y}{\sqrt{(1+y^2)}}dy=\sqrt{1+y^2}$
and also $\int \sqrt{(1+x^2)}dx$ = $\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log (x+\sqrt{1+x^2})+C$

$\therefore$ The general solution is

$\sqrt{1+y^2}$ = $\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log (x+\sqrt{1+x^2})+C$

To find the value of $C$,
use $y\left(0\right)=0$
We get, $C=1$

$\therefore$ The particular solution is

$\sqrt{1+y^2}$= $\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\log (x+\sqrt{1+x^2})+1$