- #1
Mitchtwitchita
- 190
- 0
The solution of the initial value problem, dy/dx = y-3. y(0) = 4, what does y=?
dy/y-3 = dx
1/y-3 dy = dx
ln(y-3) = ? + C
I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.
dy/y-3 = dx
1/y-3 dy = dx
ln(y-3) = ? + C
I have no idea how to do this one, can anybody please help me out? I'm having a hard time seeing how to separate the y and the -3 or if they are separable at all.