Question

What is the general solution of the differential equation `y' - (2xy)/(x^2+1) = 1`?

Answer 1

`y = (x^2+1) arctan(x) + C(x^2+1)`

Explanation:

We have:

`y' - (2xy)/(x^2+1) = 1` ..... [A}

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

`dy/dx + P(x)y=Q(x)`

As the equation is already in this form, then the integrating factor is given by;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ -(2x)/(x^2+1) \ dx)`
`\ \ = exp(- int \ (2x)/(x^2+1) \ dx)`
`\ \ = exp( - ln(x^2+1)`
`\ \ = exp( ln (1/(x^2+1) ) )`
`\ \ = 1/(x^2+1)`

And if we multiply the DE [A] by this Integrating Factor, `I`, we will have a perfect product differential;

`(1/(x^2+1)) \ y' - (1/(x^2+1)) (2xy)/(x^2+1) = (1/(x^2+1)) 1`

`:. d/dx( (1/(x^2+1)) \ y ) = 1/(x^2+1)`

Which we can directly integrate to get:

`(y)/(x^2+1) = int \ 1/(x^2+1) \ dx`

The RHS is a standard integral, so integrating we get:

`(y)/(x^2+1) = arctan(x) + C`

Leading to the General Solution:

`y = (x^2+1) arctan(x) + C(x^2+1)`