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 +ce^(-x^2)
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+ce^(-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!
dy/dx+2xy=1; y=1+ce^(-x^2); dy/dx=-2xce^(-x^2)
You can see how to use LaTeX by looking here. OP stands for "Original Post".
Your problem: verify the solution
for the DE
You'll need to use the product rule, the chain rule, and the Fundamental Theorem of the Calculus to do this verification:
You'll need to finish this computation, and then plug the result into the DE on the LHS, and see if you get the required 1 to equal the RHS. Make sense?
[EDIT] Sorry, didn't see pickslides' post.
I'm afraid it's incorrect. You can't just get rid of integrals like that. You did correctly use the FTC to simplify the first term, but the second term cannot be simplified. I think you'll find it cancels with terms in the 2xy of the original DE, as does the last term with the constant. Just leave the integrals there!