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Solution to a specific differential equation


hobz
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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.

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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.

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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. loge is often written as ln. It's just an abbreviation - they're the same thing.

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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.

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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]

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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.

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