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

Explanation

Rewrite the equation as,
`(dx)/(dy) = (tan^(-1)y)/(1+y^(2)) - (x)/(1+y^(2))`
`rArr (dx)/(dy) + (1)/(1+y^(2)) x = (tan^(-1)y)/(1+y^(2))`
This is linear in x, with from being `(dx)/(dy) + Rx = S`
IF `= e^(int Rdy) = e^(int(dy)/(1+y^(2))) = e^(tan^(-1)y)`
The solution is
`x. IF = int S. IF dy + k`, (k being the constant of integration)
`rArr x. e^(tan^(-1)y) = int e^(tan^(-1) y).(tan^(-1) y)/(1+ y^(2)) = dy` ...(i)
Put `tan^(-1)y = t rArr (dy)/(1+y^(2))= dt`
`:. int e^(tan^(-1)y)(tan^(-1)y)/(1+y^(2))dy = int t e^(1) dt = t int e^(t) dt - int e^(t) 1 dt`
`= te^(t) - e^(t) = (t-1)e^(t)`, (using Integration by parts)
From (i), `x.e^(tan^(-1)y) = (tan^(-1)y-1)e^(tan^(-1)y) + k`
`:. x = ke^(-tan^(-1)y) + tan^(-1) y- 1`