Question

Solution of the differential equation $\left(\frac{x+y-1}{x+y-2}\right) \frac{d y}{d x}=\left(\frac{x+y+1}{x+y+2}\right)$, 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 $x+y=t \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x}$.
Therefore, the given equation becomes
$\left(\frac{t-1}{t-2}\right)\left(\frac{d t}{d x}-1\right)=\frac{t+1}{t+2} $
$\Rightarrow \frac{d t}{d x}-1=\left(\frac{t+1}{t+2}\right)\left(\frac{t-2}{t-1}\right) $
$\Rightarrow \frac{d t}{d x}=\frac{t^{2}-t-2}{t^{2}+t-2}+1 $
$\Rightarrow \frac{d t}{d x}=\frac{2 t^{2}-4}{t^{2}+t-2} $
$\Rightarrow \frac{t^{2}+t-2}{t^{2}-2} d t=2 d x $
$\Rightarrow \left(1+\frac{t}{t^{2}-2}\right) d t=2 d x$
On integrating, we get
$t+\frac{1}{2} \log \left|t^{2}-2\right|=2 x+c $
$\Rightarrow (x+y)+\frac{1}{2} \log \left|(x+y)^{2}-2\right|=2 c ...$(1)
Given, $y=1$, when $x=1$, therefore $\log 2=2 c$.
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$