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
$\left(\frac{u-1}{u-2}\right)\left(\frac{du}{dx}-1\right)=\frac{u+1}{u+2}$
$\Rightarrow \frac{du}{dx}-1=\left(\frac{u+1}{u+2}\right)\left(\frac{u-2}{u-1}\right)$
$\Rightarrow \frac{du}{dx}=\frac{u^{2}-u-2}{u^{2}+u-2}+1$
$\Rightarrow \frac{u^{2}+u-2}{u^{2}-2} du=2dx$
$\Rightarrow \left(1+\frac{u}{u^{2}-2}\right)du=2dx\quad\ldots\left(i\right)$
$\Rightarrow u+\frac{1}{2}log\left|u^{2}-2\right|=2x+C\quad$ (Integrating)
$\Rightarrow \left(x+y\right)+\frac{1}{2}log\left|\left(x+y\right)^{2}-2\right|=2x+C$
$\Rightarrow 2\left(y-x\right)+log\left|\left(x+y\right)^{2}-2\right|=2C\quad\ldots\left(ii\right)$
When $x = 1$, $y = 1$, we have $log2 = 2C$
On substituting the value of $C$ in equation $(ii)$, we get
$2\left(y-x\right)+log\left|\left(x+y\right)^{2}-2\right|=log\,2$
$\Rightarrow 2\left(y-x\right)+log\left|\frac{\left(x+y\right)^{2}-2}{2}\right|=0$