Verifying solutions for differential equations

Verify this is a solution of this differential equation:

S=integrand......integrate from 0 to x

dy/dx+2xy = 1 ; y = e^(-x^2)Se^(t^2) dt +c[1]e^(-x^2)

[Difficulty]

I'm not sure how to simplify y. It looks like the fundamental theorem

of calculus. So e^(-x^2)S(from 0 to x)e^(t^2)dt should just be 1,

right? I let y=1+c[1]e^(-x^2) and used implicit differentiation, but

I end up with dy/dx+2xy=2x. If 2x=1 then I'd have it. I also

substituted y and dy/dx into the equation, but I get the same thing:

2x=1. I have to be simplifying y wrong. Can you help me? Thanks!

[Thoughts]

y=1+ce^(-x^2)

(y-1)/e^(-x^2)=c

e^(x^2)y-e^(x^2)=c

e^(x^2)y'+2xye^(x^2)-2xe^(x^2)=0

e^(x^2)y'=2xe^(x^2)-2xye^(x^2)

y'=2x-2xy

y'+2xy=2x

dy/dx+2xy=1; y=1+ce^(-x^2); dy/dx=-2xce^(-x^2)

-2xce^(-x^2)+2x(1+ce^(-x^2))=1

-2xce^(-x^2)+2x+2xce^(-x^2)=1

2x=1