Question

Find the particular solution of the differential equation $x(1+y^2)dx-y(1+x^2)dy=0$, given that $y=1$ when $x=0$.

Answer 1

$x(1+y^2)dx=y(1+x^2)dy$

The given equation is in the form of first order separable ODE has the form of $N\left(y\right)dy=M\left(x\right)dx$

Let $y$ be the dependent variable.

Divide by $dx$:

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

Rewrite in the form of first order separable ODE:

$N\left(y\right)=\frac{y}{y^2+1},M\left(x\right)=\frac{x}{x^2+1}$

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

Integrating on both the sides,

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

$\frac{1}{2}\ln (y^2+1)=\frac{1}{2}\ln (x^2+1)+C$--- (1)

When $y=1,x=0$

Equtaion (1) becomes,

$\frac{1}{2}\ln (1^2+1)=\frac{1}{2}\ln (0^2+1)+C\Rightarrow c=\frac{1}{2}\ln 2$

The required equation is $\frac{1}{2}\ln (y^2+1)=\frac{1}{2}\ln (x^2+1)+\frac{1}{2}\ln 2.$