Question

For the differential equation, find the general solution:

`x dy/dx + y - x + xy cot x = 0(x != 0)`

Answer 1

Given differential equation

`x  dy/dx + y - x + xy  cot x = 0`

⇒ `x  dy/dx + y (1 + x  cot x) = x`

or `dy/dx + (1/x + cot x) y = 1`            ...(i)

Comparing with `dy/dx + Py = Q`

`P = 1/x + cot x` and  Q = 1

∴ `I.F. = e^(int P dx) = e^(int(1/x + cot x)dx)`

`= e^(log x) + log sin x`

`=> e^(log (x sin x)) = x sin x`

Hence the required solution

∴ `y × I.F. = int I.F. xx Q  dx + C`

`=> y xx x sin x = int 1 * x sin x dx + C`

`=> xy sin x = - x cos x + int 1 cos x dx + C`

`=> xy sin x = - x cos x + sin x + C`

⇒ y = `(- x cos x)/(x sin x) + (sin x)/(x sin x) + C/(x sin x)`

⇒ `y = 1/x - cot x + C/ (x sin x)`