Question

For the differential equation given below, find a particular solution satisfying the given condition `(x + 1) dy/dx = 2e^-y + 1; y = 0` when x = 0.

Answer 1

`(x + 1) dy/dx = 2e^-y + 1; y = 0` when x = 0

`dy/dx = (2e^-y + 1)/(x + 1)`

or, `1/(2e^-y + 1) dy = 1/(x + 1) dx`

or, `e^y/(2 + e^y) dy = 1/(x + 1) dx`

Integrating both sides,

`int e^y/(2 + e^y) dy = int 1/(x + 1) dx`

or, log |2 + e^y| = log |x + 1| + log C

log |2 + e^y| = log |(x + 1)C|

2 + e^y = C(x + 1)   ...(i)

When x = 0, y = 0

2 + e⁰ = C(0 + 1)

2 + 1 = C

C = 3

Put C = 3 in equation (i)

2 + e^y = 3(x + 1)

e^y = 3x + 3 – 2

e^y = 3x + 1

`y = e^((3x  +  1))`

This is the required Particular solution.