Recent content by tiaborrego

Ok, so I think I get it now. So I solved for I I= e^{x} and thus, the general solution is: y(x)=e^x/ln[x] + c/ln[x]

So, I tried the u substitution again, and this time I got: (1/x)*e^{-1/x^2} for I. Is this right? My process: u = ln(x) du= 1/x dx I= e^{1/u*du} = e^{ln(u)*du} =e^{ln[ln(x)]*1/x} =(1/x)*e^{-1/x^2}

Ok, so I think I found the u substitution. I think it is: u= x* ln[x] du = 1+ ln[x] dx Therefore, I(x)=e^ln[x*ln(x)] = x* ln[x] Is this right?

How do you solve x ln(x) dy/dx = xe^x using the integrating factor? So far, I have put it into standard form. dy/dx + y/(xln(x))=xe^x/(x(ln(x))