No, when you multiply by the integrating factor, the RHS gets multiplied by it as well. You then have an integration by parts (on the RHS) to deal with.

Also make your mind up as to what variables you are using. Those t's should be x's.

Results 1 to 2 of 2

- Sep 15th 2009, 03:48 PM #1

- Joined
- Sep 2008
- Posts
- 11

## more IVP

y^1 (yprime) + 2y/3 = 1 - x/2 y(0) = yv0 ( y not, or y naught, or however u say it)

find y naught so that the solution just touches x-axis (can use calc)

i made it a first order ODE and made the integrating factor (u) = e ^ (int) 2dt/3

which means that....

d/dt[y(e2x/3)] = 1- x/2

so y = 9x - x^2)/(e^2x/3)

so y naught = 0.184 right?

- Sep 16th 2009, 01:17 AM #2

- Joined
- Jun 2009
- Posts
- 671
- Thanks
- 136