# Diff EQ using separation of variables

• Jan 28th 2009, 10:21 PM
Jenberl
Diff EQ using separation of variables
I am stumped on solving the following diff eq:

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

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

Please, what am I doing wrong?

Thank you,
Jen
• Jan 29th 2009, 04:49 AM
skeeter
I'll assume the $\displaystyle \frac{dy}{dx}$ in your original post was a typo ...

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

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

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

$\displaystyle y(0) = 1$ ...

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

$\displaystyle C = -\ln(3)$

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

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

$\displaystyle y = \frac{3+t}{3} = \frac{1}{3}(3 + t)$
• Jan 29th 2009, 07:28 AM
Jenberl
Thanks!
Yes, I am sorry...that WAS a typo..

Thank you so much!!

Quote:

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

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

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

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

$\displaystyle y(0) = 1$ ...

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

$\displaystyle C = -\ln(3)$

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

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

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