# Differential equations with initial conditions

• Mar 6th 2010, 08:17 PM
Quixotic
Differential equations with initial conditions
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.
• Mar 6th 2010, 08:25 PM
Prove It
Quote:

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.

$\displaystyle \frac{dy}{dt} = 2t\,e^{-2y} - e^{-2y}$ with $\displaystyle y(0) = 3$.

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

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

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

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

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

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

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

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

Now using the initial condition:

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

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

$\displaystyle 6 = \ln{C}$

$\displaystyle C = e^6$.

Therefore:

$\displaystyle y = \frac{1}{2}\ln{(2t^2 - 2t + e^6)}$.
• Mar 6th 2010, 08:30 PM
Quixotic
Quote:

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

Ah. I get it now. You have my eternal gratitude! It would have taken me -days- to work that out. (Itwasntme)