# Math Help - A simple differential equation.

1. ## A simple differential equation.

I am following an example in a book and it makes a jump that I cannot seem to follow. Please help me understand what steps were taken.

Starting Equation: $\frac{dp}{dt} = 0.5p - 450 \left(1\right)$

Which can be rewritten as:

$\frac{dp}{dt} = \frac{p-900}{2} \left(2\right)$

or if p != 900,

$\frac{\frac{dp}{dt}}{p-900} = \frac{1}{2} \left(3\right)$

Now heres where I get lost. They state that by the chain rule the left side of Eq. (3) is the derivative of $\ln |p-900|$ with respect to t, so we have:

$\frac{d}{dt} \ln |p-900| = \frac{1}{2}$

Any help is greatly appreciated.

Thanks.

2. Originally Posted by sixstringartist
Starting Equation: $\frac{dp}{dt} = 0.5p - 450 \left(1\right)$
We rewrite this one as follows

$\frac1{0,5p-450}\,dp=dt$, now integrate.

3. it seems separation of variables is the preferred method for first-order ODE's now. personally, i would have found the integrating factor, but anyway, let's go with what you have.
Originally Posted by sixstringartist
I am following an example in a book and it makes a jump that I cannot seem to follow. Please help me understand what steps were taken.

Starting Equation: $\frac{dp}{dt} = 0.5p - 450 \left(1\right)$

Which can be rewritten as:

$\frac{dp}{dt} = \frac{p-900}{2} \left(2\right)$

or if p != 900,

$\frac{\frac{dp}{dt}}{p-900} = \frac{1}{2} \left(3\right)$
ok, seems ok so far.

Now heres where I get lost. They state that by the chain rule the left side of Eq. (3) is the derivative of $\ln |p-900|$ with respect to t, so we have:

$\frac{d}{dt} \ln |p-900| = \frac{1}{2}$

Any help is greatly appreciated.

Thanks.
yes, that is a weird way to say it, but that's true. $\frac {d}{dt} \ln |p - 900|$ is indeed $\frac { dp/dt}{p - 900}$. so they just worked backwards and replaced one with the other. now they are just going to integrate both sides with respect to t.

if they are using this method, i would prefer to see this:

$\frac{\frac{dp}{dt}}{p-900} = \frac{1}{2}$ ........multiplying both sides by dt we obtain

$\frac{dp}{p-900} = \frac{1}{2}~dt$ now we integrate both sides, we get:

$\int \frac {1}{p - 900}~dp = \int \frac {1}{2}~dt$

it is often a trick used to work backwards when doing DE's, it's just something you have to rap your head around and get used to

4. it's just something you have to rap your head around and get used to
Thats proving to be a little more difficult than I had anticipated but thanks for the response.

Im new to these forums so Im still getting used to the math syntax you use to display equations. What is the notation to display a "not Equal to" sign?

5. Originally Posted by sixstringartist
Thats proving to be a little more difficult than I had anticipated but thanks for the response.

Im new to these forums so Im still getting used to the math syntax you use to display equations. What is the notation to display a "not Equal to" sign?
type \neq to obtain $\neq$

6. Originally Posted by sixstringartist
Im new to these forums so Im still getting used to the math syntax you use to display equations. What is the notation to display a "not Equal to" sign?
See the LaTeX Help

We're here to help!

P.S.: just \ne and it's done

7. Ahh yes I was hoping there was a resource for it. That saves me from having to search for the function I need. Thanks a lot.

Sorry to keep this going as I understand a method for solving now, but one thing I still dont get is how $\frac{dp/dt}{p - 900}$
is equal to

$\frac {d}{dt} \ln |p - 900|
$

by the chain rule. How does the chain rule explain this step?

8. Also you can view our LaTex Tutorial