Diff EQ using separation of variables

• Jan 28th 2009, 11:21 PM
Jenberl
Diff EQ using separation of variables
I am stumped on solving the following diff eq:

$\frac{\,dy}{\,dx}=\frac{y}{3+t}$ with the initial condition $y(0)=1$

I have figured the general solution to be $y=3+t+C$,
from $\ln\left|y\right|=\ln\left|3+t\right|+C$,
but it doesn't yield the correct result: $y=\frac{1}{3}(3+t)$
I keep getting $y=1+t$.

Please, what am I doing wrong?

Thank you,
Jen
• Jan 29th 2009, 05:49 AM
skeeter
I'll assume the $\frac{dy}{dx}$ in your original post was a typo ...

$
\frac{dy}{dt} = \frac{y}{3+t}
$

$\frac{dy}{y} = \frac{dt}{3+t}$

$
\ln|y| = \ln|3+t| + C
$

$y(0) = 1$ ...

$\ln(1) = \ln(3) + C$

$
C = -\ln(3)
$

$
\ln|y| = \ln|3+t| - \ln(3)
$

$\ln|y| = \ln\left|\frac{3+t}{3}\right|$

$
y = \frac{3+t}{3} = \frac{1}{3}(3 + t)
$
• Jan 29th 2009, 08:28 AM
Jenberl
Thanks!
Yes, I am sorry...that WAS a typo..

Thank you so much!!

Quote:

Originally Posted by skeeter
I'll assume the $\frac{dy}{dx}$ in your original post was a typo ...

$
\frac{dy}{dt} = \frac{y}{3+t}
$

$\frac{dy}{y} = \frac{dt}{3+t}$

$
\ln|y| = \ln|3+t| + C
$

$y(0) = 1$ ...

$\ln(1) = \ln(3) + C$

$
C = -\ln(3)
$

$
\ln|y| = \ln|3+t| - \ln(3)
$

$\ln|y| = \ln\left|\frac{3+t}{3}\right|$

$
y = \frac{3+t}{3} = \frac{1}{3}(3 + t)
$