# Solution to a specific differential equation

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Can someone explain to me why

$\frac{dP(t)}{P(t)}=r \, dt$

becomes

$\log |P(t)| = rt + C$

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

$\int dP(t)\frac{1}{P(t)} = \int dt \frac{dP(t)}{dt}\frac{1}{P(t)}= \ln|P(t)| + C$

More carefully;

You start from the differential equation

$\frac{dP(t)}{dt}= rP(t)$

Divide by $P(t)$ (we need to assume it is never zero)

$\frac{dP(t)}{dt}\frac{1}{P(t)}= r$

multiply by $dt$ and integrate

$\int dt \frac{dP(t)}{dt}\frac{1}{P(t)} = \int dt \:r$

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 $P(t)$ such that the derivative of it is the same function up to a factor $r$.

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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Thanks for the replies!

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

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

$\frac{\partial f(x(t))}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial f}{\partial x}$

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

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So $\frac{\partial f(x(t))}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial f}{\partial x}$ sums to one? Why?

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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 $dt$ to $dP$.

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

$t \mapsto P(t)$ (assume we can differentiate this)

then,

$dt \mapsto dP \frac{\partial P(t)}{\partial t}$

So really we should write

$\int dt \frac{\partial P(t)}{\partial t}\frac{1}{P(t)} = \int dP \frac{1}{P}= \ln |P|+C$

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

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

$\int dx f(x)$

For this situation is is the same as putting $dx$ 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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