Question

How do I solve this differential equation?

`y'=2xy+x`

Answer 1

`=> y = Ce^(x^2)-1/2`

where `C` is a arbitrary integration constant.

Explanation:

`=>y' = 2xy +x`

First, get all terms with `y` or its derivatives on one side:

`=>y' - 2xy = x`

Next, construct the integrating factor:

`=>e^(\int-2xdx)=e^(-x^2)`

Now multiply both sides of the equation by the integrating factor:

`=>e^(-x^2)y' - 2xe^(-x^2)y = xe^(-x^2)`

We can now notice that the left hand side is of the form of the product rule:

`=>d/(dx)f(x)g(x)=f(x)g'(x)+f'(x)g(x)`

So we can condense the LHS to get:

`=>d/(dx)(e^(-x^2)y) = xe^(-x^2)`

Let's move the `dx` to the RHS and integrate both sides:

`=>int d(e^(-x^2)y) = int xe^(-x^2)dx`

Integrating:

`=>e^(-x^2)y = int xe^(-x^2)dx`

We can substitute `u equiv -x^2`.
This means `du = -2xdx`

Substituting:

`=>e^(-x^2)y = -1/2 int e^u du`

`=>e^(-x^2)y = -1/2 e^u + C`

`=>e^(-x^2)y = -1/2 e^(-x^2) + C`

Solving for `y`:

`=>y = -1/2 e^(-x^2)e^(x^2) + Ce^(x^2)`

`=> y = Ce^(x^2)-1/2`

where `C` is a arbitrary integration constant.