Question

Solve the differential equation: (tan^-1y- x)dy=(1+y^2)dx

Answer 1

(Tan⁻¹ y  - x ) dy = (1+y²) dx        ---- (1)

Let Tan⁻¹ y = z     =>  y = tan z     ---- (2)
        =>    dy = Sec² z  dz

Hence,  (z - x) Sec² z dz = Sec² z  dx 
    =>   dz/dx  = 1/(z-x)       ---- (3)
Let   1/(z-x) = v    =>   z = x +1/v
   => dz/dx = 1 - 1/v²  dv/dx

Substitute in (3)....  1 - 1/v² dv/dx = v
           dv/dx = (1-v) v²
           dv / [ v² (1-v) ]  = dx
           dv [ 1/v + 1/v² + 1/(1-v) ]   = dx       by method of partial fractions
Integrate both sides.

   Ln v  -  1/v  - Ln |1-v| =  x + K
   Ln [ v/|1-v|  * exp(-1/v) ]  = x + K
   Ln [   exp(-1/v) *  1 / |1/v  - 1 |  ]   = x + K
=>   K1 * eˣ  = exp(-1/v)   * 1/|1/v  - 1|
       K * eˣ = exp(-x - tan⁻¹ y)   * [ 1/ |Tan⁻¹ y - x - 1| ]

This is the solution.