# Simple linear diff.eq.

1. Apr 11, 2009

### ImAnEngineer

Hey guys.

I've recently started studying differential equations. There is one thing I don't understand and of which I simply can't find an explanation.

I'm trying to solve some linear differential equations without using standard solutions.
Say we have the equation:
$$\frac{dp}{dt}=0.5p - 450$$

The next step is (according to my book):

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

All of the next steps that lead to the solution are clear to me. They use the chain rule to integrate, exponentiate, and get: $p=900+ce^\frac{t}{2}$.
But what I don't understand, is why they first write it in the form of eq.(1), and not as, say:
$$(2) \frac{dp}{.5p-450}=1 dt$$ ?

Last edited: Apr 12, 2009
2. Apr 11, 2009

### Pengwuino

It's equivalent, it's just that using .5 for the right hand side means the coefficient infront of p is simply 1 and the integration is slightly easier. If you used (2), a 2 would simply come out of the left hand side integration as well which makes up for the 1/2 on the right side.

Also, you do realize there is the 'dt' on the right side correct? It can't just disappear :).

3. Apr 12, 2009

### ImAnEngineer

I thought it should be equivalent, but when I integrate it I get a different solution. I'm trying to integrate it again now to see how I should get the 2 on the left hand side. Thanks for your answer.

PS: I realized that I forgot the dt's when I had turned of the computer and went to bed :D

4. Apr 12, 2009

### ImAnEngineer

Ah, now I see why the 2 comes in. If you integrate it without the 2, then you forget to compensate for the chain rule factor (or whatever you call it). I don't know if that sentence makes sense, but I get it now :) .