- #1

nominal

- 2

- 0

## 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.