Question

Solve the Differential Equation `(1+x^2)y'+4xy=(1+x^2)^-2`?

Answer 1

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

Explanation:

We have:

`(1+x^2)y'+4xy=(1+x^2)^-2` ..... [A]

We can rearrange [A] as follows:

`(1+x^2)y'+4xy = 1/(1+x^2)^2`

`dy/dx + (4x)/(1+x^2)y = 1/(1+x^2)^3` ..... [B]

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)`

So we form an Integrating Factor;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ (4x)/(1+x^2) \ dx)`
`\ \ = exp(2 \ int \ (2x)/(1+x^2) \ dx)`
`\ \ = exp(2 \ ln(1+x^2))`
`\ \ = exp(ln(1+x^2)^2)`
`\ \ = (1+x^2)^2`

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

`(1+x^2)^2 \ dy/dx + (4x(1+x^2)^2)/(1+x^2)y = (1+x^2)^2 /(1+x^2)^3`
`:. d/dx( (1+x^2)^2 \ y ) = 1 /(1+x^2)`

Which we can directly integrate to get:

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

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

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

Leading to the General Solution:

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