Question

The general solution of differential equation $x(1+y^2)dx+y(1+x^2)dy=0$ is/are:

• A. $\frac{1}{2}\ln (1+x^2)(1+y^2)=k$
• B. $(1+x^2)(1+y^2)=c$
• C. $(1+y^4)=c(1+x^2)$
• D. $\ln \left[\frac{(1+x^2)}{(1+y^2)}\right]=k$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{2}\ln (1+x^2)(1+y^2)=k$
B $(1+x^2)(1+y^2)=c$

Given equation is $x(1+y^2)dx+y(1+x^2)dy=0$

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

integrating on both sides

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

Put$1+x^2=t\Rightarrow 2xdx=dt$

and $1+y^2=u\Rightarrow 2ydy=du$

$\frac{1}{2}\int \frac{dt}{t}=-\frac{1}{2}\int \frac{du}{u}$

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

$\Rightarrow ln(1+x^2)(1+y^2)=2k$

$\Rightarrow (1+x^2)(1+y^2)=e^{2k}$