Question

Find a particular solution of the differential equation `dy/dx + y cot x = 4xcosec x(x != 0)`, given that y = 0 when `x = pi/2.`

Answer 1

Given the differential equation

`dy/dx + y cot x =  4x  cosec  x`                  ....(1)

Comparing with the linear equation `dy/dx + Py = Q`,

When  P = cot x, Q = 4x cosec x

∴ `I.F. = e^(int Pdx) = e^(int cot x  dx) = e^(log |sin x|) = sin x`

∴ The solution is `y. (I.F.) = int Q. (I.F.)  dx + C`

`therefore y sin x = int 4x  cosec x sin x dx + C`

`= int 4x dx + C = +  C`

`= (4x^2)/2 + C`

⇒ y sinx  = 2x² + C                   ....(2)

When `x = pi/2, y = 0`

∴ `0 = 2 (pi^2/4) + C`

⇒ `C = -pi^2/2`

Putting `C = pi^2/2` in (2),

`y sinx  = 2x^2 - pi^2/2 ; (sin x ne 0)`

Which is the required solution.