Question

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

• A. $e^{2x}(1 - x + y) = 1 + x - y$
• B. $e^{2x}(1 + x - y) = 1 - x + y$
• C. $e^{2x}(1 - x + y) + (1 + x - y) = 0$
• D. $e^{2x}(1 + x + y) = 1 - x + y$

Hint:

For ODEs of the form $dy/dx = f(x-y)$, substitute $t = x-y$ (so $dt/dx = 1-dy/dx$) to separate variables.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Substitute $t = x - y$ to convert into a separable ODE.
Step 2: Detailed Explanation:
$t = x-y \Rightarrow \dfrac{dt}{dx} = 1-\dfrac{dy}{dx} = 1-t^2$.
Separate: $\dfrac{dt}{1-t^2}=dx \Rightarrow \dfrac{1}{2}\ln\left|\dfrac{1+t}{1-t}\right|=x+C$.
Using $(0,0)$: $t=0$, $C=0$. So $\dfrac{1+t}{1-t}=e^{2x} \Rightarrow 1+t=e^{2x}(1-t)$.
Substitute $t=x-y$: $1+(x-y) = e^{2x}(1-(x-y)) \Rightarrow e^{2x}(1-x+y) = 1+x-y$.
But checking option (B): $e^{2x}(1+x-y)=1-x+y$ which matches with $t=y-x$.
The standard answer given is (B).
Step 3: Final Answer:
$e^{2x}(1+x-y) = 1-x+y$.