Can someone explain to me why

 

[math]\frac{dP(t)}{P(t)}=r \, dt[/math]

 

becomes

 

[math]\log |P(t)| = rt + C[/math]

 

when integrating on both sides?

 

It's the left side that is my problem. The right side is understood.

It follows from the fact that

 

[math]\int dP(t)\frac{1}{P(t)} = \int dt \frac{dP(t)}{dt}\frac{1}{P(t)}= \ln|P(t)| + C[/math]

 

More carefully;

 

You start from the differential equation

 

[math]\frac{dP(t)}{dt}= rP(t)[/math]

 

Divide by [math]P(t)[/math] (we need to assume it is never zero)

 

[math]\frac{dP(t)}{dt}\frac{1}{P(t)}= r [/math]

 

multiply by [math]dt[/math] and integrate

 

[math]\int dt \frac{dP(t)}{dt}\frac{1}{P(t)} = \int dt \:r[/math]

 

Then use my opening line to get the required result.

You should be aware of the fact that the integral of 1/x is defined as ln[x]. If you're used to integrating xⁿ to xⁿ⁺¹/n+1, for n=-1 you'll notice that the bottom of the fraction becomes 0, so the fraction as a whole is undefined, hence this method cannot be used. The integral of 1/P(t) is ln[P(t)].

 

Also, just in case, the log is in base e. log_e is often written as ln. It's just an abbreviation - they're the same thing.

In fact, to solve this you don't really need to integrate it like this. You can "spot" the solution. You are looking for a function [math]P(t)[/math] such that the derivative of it is the same function up to a factor [math]r[/math].

 

The exponential function springs to my mind!

 

You can see this is the case from you earlier work by just taking the exponential of the expression you got.

Thanks for the replies!

 

ajb, the fact you state at the beginning, is that derived from [math]\int f(t)g(t) dt[/math]?

It just comes from the function of a function rule for derivatives;

 

[math]\frac{\partial f(x(t))}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial f}{\partial x}[/math]

 

and then the integral of [math]1/x[/math], which is [math]\ln|x|+C[/math]

So [math] \frac{\partial f(x(t))}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial f}{\partial x}[/math] sums to one? Why?

There is no sum here. It is just the function of a function rule for derivatives. It allowed us to change integration from [math]dt[/math] to [math]dP[/math].

 

You could also view it as undoing the "Jacobian".

 

[math]t \mapsto P(t)[/math] (assume we can differentiate this)

 

then,

 

[math]dt \mapsto dP \frac{\partial P(t)}{\partial t}[/math]

 

So really we should write

 

[math]\int dt \frac{\partial P(t)}{\partial t}\frac{1}{P(t)} = \int dP \frac{1}{P}= \ln |P|+C[/math]

 

Note; difference between [math]d[/math] and [math]\partial[/math] and where [math]P[/math] is thought of as a function of [math]t [/math] and where it is a variable in its own right.

 

Also, don't let the (better) convention of putting [math]dx[/math] on the left confuse you, i.e.

 

[math]\int dx f(x)[/math]

 

For this situation is is the same as putting [math]dx[/math] to the right. However, sometimes it is useful to think of integration as an operator and so putting it on the left seems more natural.