
Initial Value Problem
$\displaystyle dy/dt + 0.8ty = 2t$
$\displaystyle y(0) = 9$
This is what I have so far
$\displaystyle dy/dt = 2t  0.8ty$
$\displaystyle dy/(2 0.8y) = t dt$
$\displaystyle y/2  0.8lny = (t^2)/2 + C$
Solve C:
$\displaystyle C = (9/2)0.8ln9$
Now I'm having trouble isolating Y to get an answer.
Have I taken all the right steps so far, and if so how do I go about isolating Y?
Thank You for any help

there are 2 things on this one
1. Since this is a linear DE the whole problem can be solved easier using an integrating factor
http://www.mathhelpforum.com/mathhe...ed0d1d141.gif is set up perfectly for this
with the integrating factor e^(.4*t^2)
However suppose you want to separate variables
2.first you were ok up to
http://www.mathhelpforum.com/mathhe...6ea5c47b1.gif
To integrate the left use u = 2  0.8y
you tried to write as dy/2 dy/(.8y) which is not true

Whoa I didn't not even see that it was a integrating factor problem haha thank you very much.