Question

For the differential equation, find the general solution:

(1 + x²) dy + 2xy dx = cot x dx (x ≠ 0)

Answer 1

The given equation is

(1 + x²) dy + 2xy  dx

= cot x  dx

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

Which is a liner equation of the type

Here `P = (2x)/(1 + x^2)` 

and `Q = (cot x)/(1 + x^2)`

Now `int P dx = int (2x)/(1 + x^2)  dx`

`⇒  log |1 + x^2| = log (1 + x^2)`

[∵ x² ≥ 0 ⇒ 1 + x² > 0 ⇒ |1 + x²| = 1 + x²]

∴ `I.F. = e^(int Pdx) = e^(log (1 + x^2)) = 1 + x^2`

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

⇒ `y (1 + x^2) = int cot x  dx + C`

⇒ y (1 + x²) = log |sin x| + C

⇒ y = (1 + x²)^-1 log |sin x| + C (1 + x²)^-1