- #1
MathWarrior
- 268
- 5
[itex]y' = (3+y)(1-y)[/itex]
[itex]y(0) = 2[/itex]
Attempt to solve:
[itex]\frac{dy}{dx}=(3+y)(1-y)[/itex]
[itex](3+y)(1-y)dy = dx[/itex]
[itex]\int(3+y)(1-y)dy = \int dx[/itex]
Partial Fractions:
[itex]\frac{A}{(3+y)} + \frac{B}{(1-y)}[/itex]
[itex]A(1-y)+B(3+y)[/itex]
Let y= 1 or -3
A = 1/4
B = 1/4
[itex]\frac{1}{4}\int \frac{1}{(3+y)} dy + \frac{1}{4}\int \frac{1}{(1-y)} dy[/itex]
[itex]1/4(ln(3+y)+ln(1-y)) + C = x[/itex]
Re-arranging.
[itex]ln(3+y)+ln(1-y) = 4x+C[/itex]
[itex](3+y)+(1-y) = e^{(4x+C)}[/itex]
Did I do it right so far? How do figure out y if that is the case?
[itex]y(0) = 2[/itex]
Attempt to solve:
[itex]\frac{dy}{dx}=(3+y)(1-y)[/itex]
[itex](3+y)(1-y)dy = dx[/itex]
[itex]\int(3+y)(1-y)dy = \int dx[/itex]
Partial Fractions:
[itex]\frac{A}{(3+y)} + \frac{B}{(1-y)}[/itex]
[itex]A(1-y)+B(3+y)[/itex]
Let y= 1 or -3
A = 1/4
B = 1/4
[itex]\frac{1}{4}\int \frac{1}{(3+y)} dy + \frac{1}{4}\int \frac{1}{(1-y)} dy[/itex]
[itex]1/4(ln(3+y)+ln(1-y)) + C = x[/itex]
Re-arranging.
[itex]ln(3+y)+ln(1-y) = 4x+C[/itex]
[itex](3+y)+(1-y) = e^{(4x+C)}[/itex]
Did I do it right so far? How do figure out y if that is the case?