Question

Solution of the differential equation
$\left(\frac{x+y-1}{x+y-2}\right)\frac{dy}{dx}=\left(\frac{x+y+1}{x+y+2}\right)$,
when $x=1$, $y=1$, is

• A. $log\left|\frac{{\left(x-y\right)}^2-2}{2}\right|=2\left(x+y\right)$
• B. $log\left|\frac{{\left(x-y\right)}^2+2}{2}\right|=2\left(x-y\right)$
• C. $log\left|\frac{{\left(x+y\right)}^2+2}{2}\right|=2\left(x-y\right)$
• D. none of these

Answer 1

Answer: (D)

Let $x+y=u$. Then, $1+\frac{dy}{dx}=\frac{du}{dx}$.
So, the given differential equation becomes
$\left(\frac{u-1}{u-2}\right)\left(\frac{du}{dx}-1\right)=\frac{u+1}{u+2}$
$\Rightarrow \frac{du}{dx}-1=\left(\frac{u+1}{u+2}\right)\left(\frac{u-2}{u-1}\right)$
$\Rightarrow \frac{du}{dx}=\frac{u^2-u-2}{u^2+u-2}+1$
$\Rightarrow \frac{u^2+u-2}{u^2-2}du=2dx$
$\Rightarrow \left(1+\frac{u}{u^2-2}\right)du=2dx\dots \left(i\right)$
$\Rightarrow u+\frac{1}{2}log\left|u^2-2\right|=2x+C$ (Integrating)
$\Rightarrow \left(x+y\right)+\frac{1}{2}log\left|{\left(x+y\right)}^2-2\right|=2x+C$
$\Rightarrow 2\left(y-x\right)+log\left|{\left(x+y\right)}^2-2\right|=2C\dots \left(ii\right)$
When $x=1$, $y=1$, we have $log2=2C$
On substituting the value of $C$ in equation $(ii)$, we get
$2\left(y-x\right)+log\left|{\left(x+y\right)}^2-2\right|=log2$
$\Rightarrow 2\left(y-x\right)+log\left|\frac{{\left(x+y\right)}^2-2}{2}\right|=0$