- #1
nominal
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Homework Statement
y'(t) = (y-5t)/(y+t) IVP: y(1)=0
Homework Equations
M(x,y)dx+N(x,y)dy = 0
- or do i use -
y'+p(x)y=q(x)
The Attempt at a Solution
Well I used the first equation (with M and N):
1. first checked that it was exact, by taking the partial of M and N with respect to y and t (respectively), and found that it was exact because both = 1.
2. found a solution g(x,y) = yt + (5/2)t^2 +h(y)
- solved for h(y) (by differentiating and integrating with respect to the other variable)
- and came out with g(x,y) = yt + (5/2)t^2 + y^2/2 + C
3. plugged in my initial value condition, and got C = -5/2
4. Tried plugging in for C, and then solving for y, but that didn't work out
5. Looked at the solution and wasn't even close to the answer: ln[(y 2 + 5t2 )/5] + (2/ 5) arctan[y/( 5t)] = 0
I think I'm doing the whole thing incorrectly.