# Homework Help: Rate of change

1. Jun 3, 2012

### robertjford80

1. The problem statement, all variables and given/known data

This actually part of the Friedman equation but it's the math part I'm having trouble with. I'm simplifying down to the nitty-gritty.

a'/a = x

a' = ax

a = e^x

And the lecturer said that when a rate of change is proportional to thing itself then you multiply the thing itself by exp. I don't recall learning about that rule. What is it's name so that I can go back and practice it.

Last edited: Jun 3, 2012
2. Jun 3, 2012

### LCKurtz

If $a$ is a function of $x$ so that $a' =\frac {da}{dx}$, the statement that the rate of change is proportional to itself means $\frac {da}{dx} = ka$, where $k$ is a constant. That is not the type of equation you have written.

But when you do have $\frac {da}{dx} = ka$, the solution is $a = Ce^{kx}$ for any constant $C$, as you can verify by plugging it in.

3. Jun 3, 2012

### robertjford80

but what's the name of this operation so that i can go back and look it up and practice it. the lecturer did have a constant written in the equation but I didn't write it for unknown reasons.

4. Jun 3, 2012

### HallsofIvy

The operation being used is integration and I feel sure you have heard of that before!
$$a'= \frac{da}{dx}= ax$$
so that, in terms of differentials,
$$\frac{da}{a}= xdx$$

Now integrate both sides: $ln(a)= (1/2)x^2+ C$ so that $a= e^{(1/2)x^2+ C}$ or $a(x)= C_1e^{(1/2)x^2}$ where $C_1= e^C$.

That is NOT the result LCKurtz gave because, as he said, the equation he was looking at was not the one you posted. You said "the lecturer said that when a rate of change is proportional to thing itself then you multiply the thing itself by exp." which would be the case with a'= ka, which is what LCKurtz solved, NOT with your a'= xa, which is what I solved.

a'= da/dx= ka becomes da/a= k dx which, integrated, gives ln(a)= kx+ C so that [/itex]a= C'e^{kx}[/itex] as LCKurtz said.

Last edited by a moderator: Jun 3, 2012