Question

Find a particular solution of $\frac{y}{x}\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$ , at $x=1,y=2$

Answer 1

Given

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

$\Rightarrow \int \frac{y}{1+y^2}dy=\int \frac{x}{1+x^2}dx$

$\Rightarrow \frac{1}{2}\int \frac{2y}{1+y^2}dy=\frac{1}{2}\int \frac{2x}{1+x^2}dx$

$\Rightarrow \frac{1}{2}In|1+y^2|=\frac{1}{2}In|1+x^2|+C$

Put $x=1,y=2$

$n\frac{1}{2}In5=\frac{1}{2}In2+C$

$\Rightarrow C=\frac{1}{2}ln\frac{5}{2}$

$\Rightarrow \frac{1}{2}In|1+y^2|=\frac{1}{2}In|1+x^2|+\frac{1}{2}In\frac{5}{2}$

$\Rightarrow ln(1+y^2)=ln\left(\frac{5}{2}\right(1+x^2\left)\right)$

$\Rightarrow 1+y^2=\frac{5}{2}(1+x^2)$

$\Rightarrow 2+2y^2=5+5x^2$

$\Rightarrow 5x^2-2y^2+3=0$