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## 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.