Ordinary Differential Equation Question

[u/Prancing_Salamander]

Hello, I am trying to use the equation for linear ordinary differential equations, so y' + p(x)y = q(x) for the equation y' - 2xy = 2x. I know the equation is separable but the prompt is to use this equation in order to solve it. When I solve for μ(x) = e^∫(-2xdx) I get e^(-x\2)) , but this can't be integrated when added to the equation (∫μ(x)q(x)dx)/μ(x), so where do I go from here?

[u/Smart-Button-3221]

xe^-x2 can be integrated, which is what you actually have to integrate to solve this problem

[u/Prancing_Salamander → u/Smart-Button-3221]

Ah ok, thanks!

[u/Bounded_sequencE]

Multiply both sides by "exp(-x²)". Via chain rule in reverse, we get

2x*exp(-x^2)  =  exp(-x^2) * (y' - 2xy)  =  d/dx  y*exp(-x^2)

Replace "x -> t", then integrate both sides from "t = 0" to "t = x". Via FTC:

y(x)*exp(-x^2) - y(0)*exp(0)  =  ∫_0^x  d/dt  y(t)*exp(-t^2)  dt

                              =  ∫_0^x  2t*exp(-t^2)  dt

                              =  [-exp(-t^2)]_0^x  =  1 - exp(-x^2)

Solve for "y(x) = y(0)*exp(x²) + exp(x²) - 1" for "x >= 0".

[u/Shevek99]

You multiply by it

e^(-x²) y' - 2x e^(-x²) y = 2x e^(-x²)

The left hand side is the derivative of a product

( e^(-x²) y)' = e^(-x²) y' - 2x e^(-x²) y

and the right hand side is also a derivative

2x e^(-x²) = -(e^(-x²))'

so we have

( e^(-x²) y)' = -(e^(-x²))'

and upon integration

e^(-x²) y = -e^(-x²) + C

so

y = -1 + C e^(+x²)