Originally Posted by

**Quixotic** I have a practice problem that confuses me.

$\displaystyle y'= 2te^{-2y} -e^{-2y} $ y(0)=3

I worked this out as

$\displaystyle \int e^{2y} \frac {dy}{dt} dt= \int 2t -1 dt $

then

$\displaystyle e^{2y} = 2t^2 -2t +2C

$

$\displaystyle

2y= ln(2t^2 -2t + 2C)

$

$\displaystyle

y = \frac{1}{2} ln(2t^2 -2t + 2C)

$

when y(0)=3 is used, the worksheet and I disagree on the outcome.

$\displaystyle

3 = y(0) =\frac{1}{2} ln(2t^2 -2t + 2C)

$

$\displaystyle

3 = \frac{1}{2} ln(2(0)^2 -2(0) + 2C)

$

$\displaystyle

3 = \frac{1}{2} ln 2C

$

$\displaystyle

e^6 = 2C

$

$\displaystyle

\frac{e^6}{2}

$

Which would result in a final equation looking like:

$\displaystyle

y = \frac{1}{2} ln(2t^2 -2t + \frac{e^6}{2})

$

But my worksheet declares the correct answer to be:

$\displaystyle

y = \frac{1}{2} ln(2t^2 -2t + e^6)

$

How did I end up with the extra $\displaystyle \frac{1}{2} $ ?

I'm quite confused, grateful for any explanation, thanks.