# Thread: Differential equations with initial conditions

1. ## Differential equations with initial conditions

I have a practice problem that confuses me.

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

I worked this out as

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

then

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

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

$

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

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

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

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

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

$
e^6 = 2C
$

$
\frac{e^6}{2}
$

Which would result in a final equation looking like:
$

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

But my worksheet declares the correct answer to be:
$
y = \frac{1}{2} ln(2t^2 -2t + e^6)
$

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

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

2. Originally Posted by Quixotic
I have a practice problem that confuses me.

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

I worked this out as

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

then

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

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

$

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

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

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

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

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

$
e^6 = 2C
$

$
\frac{e^6}{2}
$

Which would result in a final equation looking like:
$

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

But my worksheet declares the correct answer to be:
$
y = \frac{1}{2} ln(2t^2 -2t + e^6)
$

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

I'm quite confused, grateful for any explanation, thanks.
$\frac{dy}{dt} = 2t\,e^{-2y} - e^{-2y}$ with $y(0) = 3$.

$\frac{dy}{dt} = e^{-2y}(2t - 1)$

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

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

$\int{e^{2y}\,dy} = t^2 - t + C_1$

$\frac{1}{2}e^{2y} + C_2 = t^2 - t + C_1$

$e^{2y} = 2t^2 - 2t + C$, where $C = 2(C_1 - C_2)$

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

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

Now using the initial condition:

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

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

$6 = \ln{C}$

$C = e^6$.

Therefore:

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

3. $e^{2y} = 2t^2 - 2t + C$, where $C = 2(C_1 - C_2)$
Ah. I get it now. You have my eternal gratitude! It would have taken me -days- to work that out.