Question

ODE: Solve the initial value problem:
3y'' - y' + (x +1)y = 1 with y(0) = y'(0) = 0

Answer 1

the roots of:

3r^2 -r + (x+1) = 0 might be a good start

Answer 2

\[roots=\frac{1\pm \sqrt{-12x-11}}{6}\]
but iffy if that helps

Answer 3

\[y=c_1\ e,(\frac{1+\sqrt{-12x-11}}{6})+C_2\ e,(\frac{1+\sqrt{-12x-11}}{6})\] if the roots are to be real; which im leary about

Answer 4

taking diffy qs in teh spring; hopefully ill be better at them by then :)

Answer 5

the wolf gives something ugly:


http://www.wolframalpha.com/input/?i=3y%27%27+-+y%27+%2B+%28x+%2B1%29y+%3D+1%2C+y%280%29+%3D+0%2C+y%27%280%29+%3D+0

Answer 6

thanks!