Question

Find particular solution of different equation $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$ given that x = 0 y = 1

Answer 1

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

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

${\tan }^{-1}y+C_1={\tan }^{-1}x+C_2$

$\Rightarrow {\tan }^{-1}x-{\tan }^{-1}y+C=0.....\left(1\right)$

Given that:- $x=0,y=1$

${\tan }^{-1}\left(0\right)-{\tan }^{-1}\left(1\right)+C=0$

$\Rightarrow C=\frac{\pi }{4}$

Substituting the value of $x$ in equation $\left(1\right)$, we get

${\tan }^{-1}x-{\tan }^{-1}y+\frac{\pi }{4}=0$

Hence the particular solution of the given differential equation is ${\tan }^{-1}x-{\tan }^{-1}y+\frac{\pi }{4}=0$.