The problem statement, all variables and given/known data dy/dx + y/x = e^(x^2) Express y in terms of x and arbitrary constant. The attempt at a solution It is in the standard 1st order linear ODE form. P(x) = 1/x Q(x) = e^(x^2) u(x) = x (after calculation) So, d(uy)/dx = uQ d(uy)/dx = x.e^(x^2) I have to integrate both sides w.r.t.x Finding the R.H.S is problematic though. As it seems infinite, from what i've understood from my calculations. Integral of x.e^(x^2) (done by partial integration) Let U = x, so dU/dx = 1 Let dV = e^(x^2), so V = [e^(x^2)]/2x (is this correct?) xy = x.[e^(x^2)]/2x - integral of [e^(x^2)]/2x.(1).dx Then i have to integrate the R.H.S again and again and again, as i can't get rid of e^(x^2) with a multiple of x always in the denominator. Any advice?
Hi Quinzio I know about that formula. I used it in the first place, and the spot where i'm stuck is due to b(x) in your equation being e^(x^2). The power of e contains x which is giving me quite some trouble with this problem.
Ok, but you got an x which is its derivative (of [itex]x^2[/itex]). It goes: [tex]a(x)=1/x[/tex] [tex]\int a(x)dx=ln|x|[/tex] [tex]e^{\int a(x)dx}=|x|[/tex] [tex]\int e^{\int a(x)dx}b(x)dx=|x|e^{x^2}=\frac{1}{2}e^{x^2}*sgn(x)[/tex]
I did some research and it appears that the integral of e^(x^2) is infinity and cannot be expressed in normal terms. However, it is curious that my calculus notes give this as the answer to this problem: y= [e^(x^2)]/2x + A/x
Your notes show the right answer. I just wrote a part of it, what I thought was the harder. The answer in your notes is for x>=0 only, anyway.
This is the last line where i've been able to reach: xy = x.[e^(x^2)]/2x - integral of [e^(x^2)]/2x.(1).dx I'm stuck at the R.H.S. where i need to integrate [e^(x^2)]/2x I realize that i know nothing about the sign function. If that is a requirement to solve this problem, then i won't be able to do this. Is there any easy guide that you could possibly recommend to learn about the sign function?