Question

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

• A. $-{\log }_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$
• B. ${\log }_e\left|\frac{2-y}{2-x}\right|=2(y-1)$
• C. ${\log }_e\left|\frac{2-x}{2-y}\right|=x-y$
• D. $-{\log }_e\left|\frac{1+x-y}{1-x+y}\right|=x+y-2$

Answer 1

The correct option is A $-{\log }_e\left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$

$\frac{dy}{dx}=(x-y{)}^2$
Put $x-y=v$
$\Rightarrow 1-\frac{dy}{dx}=\frac{dv}{dx}\cdots \left(1\right)$
$\therefore \frac{dy}{dx}=v^2\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow 1-v^2=\frac{dv}{dx}$
$\Rightarrow \frac{dv}{1-v^2}=dx$
Integrating both sides
$\Rightarrow \int \frac{dv}{1-v^2}=\int dx$

$\Rightarrow \frac{1}{2}\log \left|\frac{1+v}{1-v}\right|=x+c$

$\Rightarrow \frac{1}{2}\log \left|\frac{1+x-y}{1-x+y}\right|=x+c$

At $x=1,y=1$
$\Rightarrow \frac{1}{2}\log 1=1+c\Rightarrow c=-1$

$\Rightarrow \frac{1}{2}\log \left|\frac{1+x-y}{1-x+y}\right|=x-1$

$\Rightarrow -\log \left|\frac{1-x+y}{1+x-y}\right|=2(x-1)$