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

Explanation

The given differential equation may be written as
`(dy)/(dx) + (1)/(( 1+ x ^(2))) * y = (tan ^(-1)x)/(( 1+ x ^(2)))`
This is of the form ` (dy )/(dx) + Py =Q`, where `P = (1)/(( 1+ x ^(2))) and Q = (tan ^(-1) x )/( (1 + x^(2)))`.
Thus, the given equation is linear.
` IF= e^(int Px) = e ^(int (1)/(1 + x^(2)) dx) = e ^( tan ^(-1)x)`
` therefore ` the required solution is
` y xx IF = int {Q xx IF} dx + C`,
i.e., `y xx e ^(tan ^(-1)x) = int { (tan ^(-1)x)/((1 + x ^(2))) * e ^(tan ^(-1)x)} dx + C `
` " " = int (t e^(t)) dt + C, `where `tan ^(-1) x = t `
` " " = t e ^(t) - int 1 * e^(t) dt + C ` [integrating by parts]
`" "= t e ^(t) - e ^(t) + C = e ^(t) ( t- 1 ) +C `
` " " = e ^(tan ^(-1)x) (tan ^(-1)x - 1 ) + C`
`rArr y = (tan ^(-1)x- 1 )+ Ce^(-tan ^(-1)x) `
Hence, ` y = (tan ^(-1) x -1 ) + C e ^(-tan ^(-1)x) ` is the required solution.