superbeer

Thanks for the reply The problem is the you can't take the natural logarithm of a negative number.

Hi all, I'm trying to solve this differential equation but I'm getting strange results. Don't know how type all the symbols so I hope it makes sense. dy/dx + 3y = 6 dy/dx = 6 - 3y 1/(6 - 3y)(dy/dx) = 1 1/(6 -3y)dy = dx if I integrate both sides (-1/3 )ln[6-3y] = x + C boundary cond. when y=3 x=0 to find C, I substitute (-1/3)ln[6-3(3)] = C ok, this is where I'm stuck. Do I take C to be zero in this case. Also, does anyone know of a website where I can get a simple and easy introduction to first order linear differential equations? Thanks