Question

For the following differential equation find the particular solution satisfying the given condition:

`(e^y + 1) cos x + e^y sin x. dy/dx = 0,  "when" x = pi/6,` y = 0

Answer 1

`(e^y + 1) cos x + e^y sin x dy/dx = 0`

`e^y.sinx.dy/dx = - (e^y + 1) cosx`

`inte^y/(e^y + 1).dy = - intcosx/sinx. dx`

`log |e^y + 1| = - log |sinx| + log |c|`

`log |e^y + 1| + log |sinx| = log|c|`

`log|(e^y + 1) . sinx| = log |c|`

`(e^y + 1). sinx = c` ...(i)

when `x = pi/6, y = 0`

`(e^0 + 1). sin(pi/6) = 0`

`(1 + 1) . 1/2 = c`

`2 xx 1/2 = c`

c = 1

From (i)

∴ the particular solution is (e^y + 1). sinx = 1