Question

Solution of the differential equation $\left(\frac{x+y-1}{x+y-2}\right)\frac{dy}{dx}=\left(\frac{x+y+1}{x+y+2}\right)$, given that $y=1$ when $x=1$, is $k(y-x)+\log \left|\frac{(x+y{)}^k-k}{k}\right|=0$, where $k=$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let $x+y=t\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$.
Therefore, the given equation becomes
$\left(\frac{t-1}{t-2}\right)\left(\frac{dt}{dx}-1\right)=\frac{t+1}{t+2}$
$\Rightarrow \frac{dt}{dx}-1=\left(\frac{t+1}{t+2}\right)\left(\frac{t-2}{t-1}\right)$
$\Rightarrow \frac{dt}{dx}=\frac{t^2-t-2}{t^2+t-2}+1$
$\Rightarrow \frac{dt}{dx}=\frac{2t^2-4}{t^2+t-2}$
$\Rightarrow \frac{t^2+t-2}{t^2-2}dt=2dx$
$\Rightarrow \left(1+\frac{t}{t^2-2}\right)dt=2dx$
On integrating, we get
$t+\frac{1}{2}\log \left|t^2-2\right|=2x+c$
$\Rightarrow (x+y)+\frac{1}{2}\log \left|(x+y{)}^2-2\right|=2c...$(1)
Given, $y=1$, when $x=1$, therefore $\log 2=2c$.
Substituting the value of $c$ in (1), we get
$2(y-x)+\log \left|(x+y{)}^2-2\right|=\log 2$
$\Rightarrow 2(y-x)+\log \left|\frac{(x+y{)}^2-2}{2}\right|=0$.
$\therefore k=2$