
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.

So you have
$\displaystyle w\dfrac{u''(w)}{u'(w)}=r,$ where $\displaystyle r>0$ is a constant. Correct?


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
$\displaystyle w\dfrac{z'(w)}{z(w)}=r.$ You can divide both sides by $\displaystyle w$ to obtain
$\displaystyle \dfrac{z'(w)}{z(w)}=\dfrac{r}{w}.$
You can now integrate both sides directly. You obtain
$\displaystyle \lnz(w)=r\lnw.$ Can you continue from here?

do i also have to multiply each side by d/dw ?

No. Right now, I would solve for z(w), and then plug back in what that was. What do you get?

z(w) = e ^ r ln(w)
u(w) = the above integrated
i wish i knew how to type this math notation...

You can simplify the above expression a good deal:
$\displaystyle z(w)=e^{r\lnw}=e^{\ln(w^{r})}=(e^{\lnw})^{r}=w^{r}.$
Incidentally, I forgot a constant of integration. We would get that back at this step:
$\displaystyle \lnz(w)=r\lnw+C,$ which would translate to the multiplication of an arbitrary constant (equal to $\displaystyle e^{C}$), which gives us
$\displaystyle z(w)=Aw^{r}.$ Now continue?
Incidentally, you can learn something about LaTeX here.