Question

Equation of the curve passing through (0, 0) and satisfying the equation $\frac{dy}{dx}=(x-y{)}^2$

• 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

Answer: (A)

$e^{2x}(1-x+y)=1+x-y$

$\frac{dy}{dx}=(x-y{)}^2$ ...... $\left(i\right)$

Put $x-y=v$

$\Rightarrow 1-\frac{dv}{dx}=\frac{dy}{dx}$

$\Rightarrow 1-\frac{dv}{dx}=v^2$ ......... [From $\left(i\right)$]

$\Rightarrow dx=\frac{dv}{(1-v)(1+v)}$

$\Rightarrow \int 2dx=\int \frac{1}{1+v}+\frac{1}{1-v}dv$

$\Rightarrow 2x+c=\log (1+v)-\log (1-v)$

$\Rightarrow 2x+c=\log \left({(1-v)}^{-1}(1+v)\right)$

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

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

Put $(x,y)=(0,0)$

$\Rightarrow 1=1.e^c$

$\Rightarrow e^c=1$

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