Question

For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1

Answer 1

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

∴ `(x + 1) dy /dx = 2/e^y + 1`

∴ `(x + 1) dy /dx = ((2+e^y))/e^y `

∴ `e^y /(2+e^y) dy= dx/(1+x)`

Integrating on both sides, we get

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

∴ log| 2 + e^y| = log |1 + x| + log |c|

∴ log |2 + e^y| = log |c(1 + x)|

∴ 2 + e^y = c (1 + x)         ...(i)

When y = 0, x = 1, we have

2 + e⁰ = c (1 + 1)

∴ 2 + 1 = 2c

∴ c = `3/2`

Substituting c = `3/2` in (i), we get

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

∴ 4 + 2e^y = 3 + 3x

∴  3x - 2e^y - 1 = 0, which is the required particular solution.