The solution of the differential equation \((x + 1)\displaystyle\frac{dy}{dx} + 1 - 2e^{-y} = 0;\ y(0) = 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(|(x+1)(e^y-2)| = 1\)
The equation is in variables separable form:
\( \displaystyle (x + 1)\frac{dy}{dx} = 2e^{-y} - 1 \)
\( \displaystyle \frac{dy}{2e^{-y} - 1} = \frac{dx}{x + 1} \)
Multiplying the numerator and denominator on the left by \(e^y\):
\( \displaystyle \frac{e^y\,dy}{2 - e^y} = \frac{dx}{x + 1} \)
Integrating both sides (the left numerator is minus the derivative of \(2 - e^y\)):
\(-\log|2 - e^y| = \log|x + 1| + C\)
\(\log|(x + 1)(2 - e^y)| = -C\)
\(|(x + 1)(e^y - 2)| = K,\ \text{where } K = e^{-C}\)
Putting x = 0, y = 0:
\(|(1)(1 - 2)| = K,\quad K = 1\)
So the solution is \(|(x+1)(e^y-2)| = 1\).
Tip: The initial condition alone does not decide this question: Option (1) also gives \(|(-1)(-1)| = 1\) at (0, 0). Only the differential equation shows that the factor is \(x + 1\).