# Differential equation question

• April 19th 2008, 10:58 AM
emttim84
Differential equation question
Here's the instructions:

In Exercises 37-39, use Newton's Law of Cooling, which states that the rate of change in the temperature T of an object is proportional to the difference between the temperature T of the object and the temperature T0 of the surrounding environment. This is described by the differential equation $\frac{dT}{dt}=k(T-T_0)$

In the solutions manual they alter the DE to $T=Ce^{kt}+T_0$

How do they change it like that? The section it's in is separation of variables so I've been getting all the x's on one side, all the y's on one side, etc. but I don't quite see how they do that at all in this instance...
• April 19th 2008, 11:16 AM
Peritus
$
\frac{1}
{k}\frac{{dT}}
{{T - T_0 }} = dt
$

now integrate both sides:

$
\begin{gathered}
\frac{1}
{k}\ln \left( {T - T_0 } \right) = t + \beta \hfill \\
\hfill \\
\leftrightarrow T - T_0 = e^{kt + k\beta } \hfill \\
\hfill \\
\leftrightarrow T(t) = \underbrace {e^{k\beta } }_Ce^{kt} + T_0 = Ce^{kt} + T_0 \hfill \\
\end{gathered}
$
• April 19th 2008, 11:28 AM
emttim84
Quote:

Originally Posted by Peritus
$
\frac{1}
{k}\frac{{dT}}
{{T - T_0 }} = dt
$

now integrate both sides:

$
\begin{gathered}
\frac{1}
{k}\ln \left( {T - T_0 } \right) = t + \beta \hfill \\
\hfill \\
\leftrightarrow T - T_0 = e^{kt + k\beta } \hfill \\
\hfill \\
\leftrightarrow T(t) = \underbrace {e^{k\beta } }_Ce^{kt} + T_0 = Ce^{kt} + T_0 \hfill \\
\end{gathered}
$

You are a lifesaver, thank you so much!
• April 19th 2008, 12:20 PM
colby2152
Quote:

Originally Posted by emttim84
You are a lifesaver, thank you so much!

Tim, I believe you are much more of a lifesaver i.e. your signature being true!(Handshake)