Question

Find the particular solution of the following differential equation, given that y = 0 when x = `pi/4`.

`(dy)/(dx) + ycotx = 2/(1 + sinx)`

Answer 1

The differential equation is a linear differential equation

IF = `e^(int cotxdx) = e^(logsinx) = sinx`

The general solution is given by

`ysinx = int 2 sinx/(1 + sinx) dx`

⇒ `ysinx = 2 int (sinx + 1 - 1)/(1 + sinx) dx = 2 int [1 - 1/(1 + sinx)] dx`

⇒ `ysinx = 2 int [1 - 1/(1 + cos(pi/2 - x))] dx`

⇒ `ysinx = 2 int [1 - 1/(2cos^2 (pi/4 - x/2))] dx`

⇒ `ysinx = 2 int [1 - 1/2 sec^2 (pi/4 - x/2)] dx`

⇒ `ysinx = 2[x + tan(pi/4 - x/2)] + c`

Given that y = 0, when x = `pi/4`,

Hence, 0 = `2[pi/4 + tan  pi/8] + c`

⇒ `c = - pi/2 - 2 tan  pi/8`

Hence, the particular solution is `y = "cosec"x [2{x + tan  (pi/4 - x/2)} - (pi/2 + 2tan  pi/8)]`