- #1
cklabyrinth
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Hello. I need some help solving a differential equation. I think where I'm going wrong is integrating one side via partial fractions, but I'm not quite sure where my mistake is. Using Wolfram, I found the correct answer, which is below. Thanks.
Solve the following initial value problem:
[itex]ty' = y^2 - y[/itex]
[itex]y(1) = \frac{1}{2}[/itex]
[itex]y = ?[/itex]
n/a
[itex]\int \frac{dy}{y(y-1)}\ [/itex] Eq. 1
[itex]\frac{1}{y(y-1)} = \frac{A}{y} + \frac{B}{y-1} [/itex]
[itex]A(y-1) + By = 1 [/itex]
[itex]y = 1 -> B = 1 [/itex]
[itex]y = 0 -> A = -1 [/itex]
So:
[itex]\int -\frac{1}{y} + \frac{1}{y-1}dy = ln|y-1|-ln|y| + c = ln|\frac{y-1}{y}| + c[/itex]
However, Wolfram tells me that the answer is actually:
[itex]ln|\frac{1-y}{y}| + c[/itex]
If I go on to use that in the problem, I get the correct answer for the DE.
[itex]y = \frac{1}{t + 1} (c = 0)[/itex]
With my answer, I get the incorrect overall answer of:
[itex]y = \frac{1}{1 - t}[/itex]
How am I solving the partial fractions problem incorrectly?
Thanks.
Homework Statement
Solve the following initial value problem:
[itex]ty' = y^2 - y[/itex]
[itex]y(1) = \frac{1}{2}[/itex]
[itex]y = ?[/itex]
Homework Equations
n/a
The Attempt at a Solution
[itex]\int \frac{dy}{y(y-1)}\ [/itex] Eq. 1
[itex]\frac{1}{y(y-1)} = \frac{A}{y} + \frac{B}{y-1} [/itex]
[itex]A(y-1) + By = 1 [/itex]
[itex]y = 1 -> B = 1 [/itex]
[itex]y = 0 -> A = -1 [/itex]
So:
[itex]\int -\frac{1}{y} + \frac{1}{y-1}dy = ln|y-1|-ln|y| + c = ln|\frac{y-1}{y}| + c[/itex]
However, Wolfram tells me that the answer is actually:
[itex]ln|\frac{1-y}{y}| + c[/itex]
If I go on to use that in the problem, I get the correct answer for the DE.
[itex]y = \frac{1}{t + 1} (c = 0)[/itex]
With my answer, I get the incorrect overall answer of:
[itex]y = \frac{1}{1 - t}[/itex]
How am I solving the partial fractions problem incorrectly?
Thanks.