Question

What is a solution to the differential equation `dy/dx=1+2xy`?

Answer 1

`y = sqrt(pi)/2 e^(x^2) "erf"(x) + Ae^(x^2)`

Where `"erf"(x)` is the Error Function :

`"erf"(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt`

Explanation:

`dy/dx = 1 + 2xy`
`:. dy/dx - 2xy = 1` ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

`I = e^(int P(x) dx)`
`\ \ = e^(int \ -2x \ dx)`
`\ \ = e^(-x^2)`

And if we multiply the DE [1] by this Integrating Factor we will have a perfect product differential;

`dy/dx - 2xy = 1`
`:. e^(-x^2)dy/dx - 2xye^(-x^2) = 1*e^(-x^2)`
`:. d/dx(ye^(-x^2)) = e^(-x^2)`

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

`ye^(-x^2) = int \ e^(-x^2) \ dx`

The RHS integral does not have an elementary form, but we can use the definition of the Error Function :

`"erf"(x) = 2/sqrt(pi) int_0^x e^(-t^2) \ dt`

Which gives us:

`ye^(-x^2) = sqrt(pi)/2"erf"(x) + A`
`y = sqrt(pi)/2 e^(x^2) "erf"(x) + Ae^(x^2)`