Question

The equation of the curve passing through the origin and satisfying the differential equation $\frac{dy}{dx}=(x-y{)}^2$ is:

• A. $e^{2x}(1-x+y)=1+x-y$
• B. $e^{2x}(1+x-y)=1-x+y$
• C. $e^{2x}(1+x-y)=-(1+x+y)$
• D. $e^{2x}(1+x+y)=1-x+y$

Answer 1

The correct option is A $e^{2x}(1-x+y)=1+x-y$

Let $x-y=v$
$\Rightarrow \frac{dy}{dx}=1-\frac{dv}{dx}$
$\Rightarrow 1-\frac{dv}{dx}=v^2$
$\Rightarrow \frac{dv}{(1-v^2)}=dx$
$\Rightarrow \int \frac{1}{2}(\frac{1}{1+v}+\frac{1}{1-v})dv=\int dx+C$
$\Rightarrow \frac{1}{2}\log (1+v)-\log (1-v)=x+C$
$\Rightarrow \log \left(\frac{1+v}{1-v}\right)=2x+c$ $(2C=c)$
$\Rightarrow (1+x-y)=e^{2x+c}(1-x+y)$

$(x,y)=(0,0)$
$\Rightarrow 1=e^c$ $\Rightarrow c=0$
$\Rightarrow (1+x-y)=e^{2x}(1-x+y)$

Hence, option A.