Solve (x + 1) dy/dx= 2e^-y - 1, given that x = 0 when y = 0. ​

Question

Solve (x + 1) \(\frac{dy}{dx}\) = 2e^-y - 1, given that x = 0 when y = 0.

Answer 1

Given Equation: (x + 1) \(\frac{dy}{dx}\) = 2e^-y -1

Re-arranging, we get,

Integrating both sides, we get,

log t = log(x + 1) + C

log (2 – e^y) = log (x + 1) + C

At x = 0, y = 0.

Therefore,

log(2) = log(1) + C

Therefore,

C = log 2

Now, we have,

log (2 – e^y) – log (x + 1) – log 2 = 0