Question

Solve the following differential equation.

$\left(\frac{x+y-1}{x+y-2}\right)\frac{dy}{dx}=\frac{x+y+1}{x+y+2}$

Answer 1

Consider the given differential equation.

$\left(\frac{x+y-1}{x+y-2}\right)\frac{dy}{dx}=\frac{x+y+1}{x+y+2}$

Let,

$u=x+y$

$\frac{du}{dx}=1+\frac{dy}{dx}$

Therefore,

$\left(\frac{x+y-1}{x+y-2}\right)\frac{dy}{dx}=\frac{x+y+1}{x+y+2}$

$\frac{u-1}{u-2}(\frac{du}{dx}-1)=\frac{u+1}{u+2}$

$\left(\frac{u-1}{u-2}\right)\frac{du}{dx}=\frac{u+1}{u+2}+\frac{u-1}{u-2}$

$\left(\frac{u-1}{u-2}\right)\frac{du}{dx}=\frac{u^2+u-2u-2+u^2-u+2u-2}{(u-2)(u+2)}$

$\left(\frac{u-1}{u-2}\right)\frac{du}{dx}=\frac{2(u^2-2)}{(u-2)(u+2)}$

$\frac{u^2-4}{u^2-2}du=2dx$

Integrate both the sides.

$\int \frac{u^2-4}{u^2-2}du=\int 2dx$

$u+\frac{\ln \left(\frac{|u+\sqrt{2}|}{|u-\sqrt{2}|}\right)}{\sqrt{2}}=2x+c$

$x+y+\frac{1}{\sqrt{2}}\ln \left(\frac{|x+y+\sqrt{2}|}{|x+y-\sqrt{2}|}\right)=2x+c$

$x+y+\frac{1}{\sqrt{2}}\ln \left(\frac{|x+y+\sqrt{2}|}{|x+y-\sqrt{2}|}\right)+C=0$

Hence, this is the required solution of the integral.