# Economics Problem

• Dec 8th 2010, 10:22 AM
ktcyper03
Economics Problem
So I've never taken differential equations, but this problem popped up in an Econ class so I'm pretty lost.

Let R(w) = {(-w)[u''(w)]}/u'(w)

Suppose R(w) = r > 0 for all w.

Derive an expression for u(w).

So i began by multiplying both sides by u'(w) and then bringing everything over to the left. I'm not sure where to introduce d/dw. I guess I'm pretty lost. There was a hint: let z(w) = u'(w) and solve the differential equation for z.

This then becomes:
R(w) = {(-w)[z'(w)]}/z(w)
[z(w)]R(w) = -w[z'(w)]
[z(w)]R(w) + w[z'(w)] = 0

Then I'm stuck.
Thanks in advance for the help.
• Dec 8th 2010, 10:41 AM
Ackbeet
So you have

$-w\dfrac{u''(w)}{u'(w)}=r,$ where $r>0$ is a constant. Correct?
• Dec 9th 2010, 11:54 AM
ktcyper03
correct!!
• Dec 9th 2010, 11:58 AM
Ackbeet
Right then. I wouldn't use the R(w) notation, because that's essentially an extra variable. Follow the hint (which reduces the order of the DE by one), and you get

$-w\dfrac{z'(w)}{z(w)}=r.$ You can divide both sides by $-w$ to obtain

$\dfrac{z'(w)}{z(w)}=-\dfrac{r}{w}.$

You can now integrate both sides directly. You obtain

$\ln|z(w)|=-r\ln|w|.$ Can you continue from here?
• Dec 9th 2010, 12:20 PM
ktcyper03
do i also have to multiply each side by d/dw ?
• Dec 9th 2010, 12:22 PM
Ackbeet
No. Right now, I would solve for z(w), and then plug back in what that was. What do you get?
• Dec 9th 2010, 12:32 PM
ktcyper03
z(w) = e ^ -r ln(w)

u(w) = the above integrated

i wish i knew how to type this math notation...
• Dec 9th 2010, 12:57 PM
Ackbeet
You can simplify the above expression a good deal:

$z(w)=e^{-r\ln|w|}=e^{\ln(|w|^{-r})}=(e^{\ln|w|})^{-r}=w^{-r}.$

Incidentally, I forgot a constant of integration. We would get that back at this step:

$\ln|z(w)|=-r\ln|w|+C,$ which would translate to the multiplication of an arbitrary constant (equal to $e^{C}$), which gives us

$z(w)=Aw^{-r}.$ Now continue?

Incidentally, you can learn something about LaTeX here.