Question

Find a particular solution of the differential equation`(x + 1) dy/dx = 2e^(-y) - 1`, given that y = 0 when x = 0.

Answer 1

The given equation is

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

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

Integrating, we get `int dy/(2e^-y - 1) = int dx/(x + 1) + C`

⇒ `int dy/(2e^-y - 1) =log |x + 1| + C`

Now, `I = int dy/ (2e^-y - 1) = int e^y/(2 - e^y)  dy`

Put e^y = t so that e^y dy = dt

∴ `I = int dt/(2-t) = - log |2 - t| = - log |2 - e^y|`

From (1), - log |2 - e^y|

= log |x + 1| + C                  ....(2)

When x = 0. y = 0

∴ - log |2 - 1| = log |0 + 1| + C

⇒ - log |1| = log |1| + C

⇒ 0 = 0 + C

⇒ C = 0

Putting in (2), - log |2 e^y| = log |x + 1|

⇒ `log |2 - e^y| = log |1/ (x + 1)|`

⇒ `2 e^y = 1/(x + 1)`

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

⇒ `y = log |(2x + 1)/ (x + 1)|, x ne -1`

Which is the required solution.