# Initial Value Problem

• April 29th 2009, 09:32 PM
Krooger
Initial Value Problem
$dy/dt + 0.8ty = 2t$
$y(0) = 9$
This is what I have so far
$dy/dt = 2t - 0.8ty$
$dy/(2- 0.8y) = t dt$
$y/2 - 0.8ln|y| = (t^2)/2 + C$
Solve C:
$C = (9/2)-0.8ln|9|$

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
• April 30th 2009, 12:05 AM
Calculus26
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/math-he...ed0d1d14-1.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/math-he...6ea5c47b-1.gif

To integrate the left use u = 2 - 0.8y

you tried to write as dy/2 -dy/(.8y) which is not true
• April 30th 2009, 01:57 AM
Krooger
Whoa I didn't not even see that it was a integrating factor problem haha thank you very much.