Question

The equation of the curve passing through the origin and satisfying the differential 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. $e^{2x}(1-x+y) = -(1+x+y)$

Hint:

Equations of form $y'=f(ax+by+c)$ are always solved using the substitution $v = ax+by+c$.

Answer 1

The Correct Option is A

Step 1: Concept
Use substitution $v = x-y$ to simplify the differential equation.

Step 2: Meaning
$v = x-y \implies \frac{dv}{dx} = 1 - \frac{dy}{dx} \implies \frac{dy}{dx} = 1 - \frac{dv}{dx}$. Substituting into the equation: $1 - \frac{dv}{dx} = v^2 \implies \frac{dv}{dx} = 1 - v^2$.

Step 3: Analysis
Separate and integrate: $\int \frac{dv}{1-v^2} = \int dx \implies \frac{1}{2} \log(\frac{1+v}{1-v}) = x + c$. Curve passes through (0,0), so $x=0, y=0 \implies v=0$. This gives $c=0$.

Step 4: Conclusion
$\log(\frac{1+x-y}{1-x+y}) = 2x \implies \frac{1+x-y}{1-x+y} = e^{2x} \implies 1+x-y = e^{2x}(1-x+y)$. Final Answer: (A)