I am currently reading "The Economics of Superstar" written by Rosen (1981). I don't understand one differential equation he used in the paper. The equation is as follows: $$\frac{dp}{dz} = (p+s)/z.$$

He says if we integrate this, we get $p(z) = vz-s$ for $v = (p+s)/z$. I think that it is a simple differential equation, but I am having trouble solving this. Could you give some help?

  • 2
    $\begingroup$ Hi: There may be an easier way (my knowledge is that I took a course in dfeqs 30 years ago ) but, if you re-write it as $dp/dz - p/z = s/z. $, then, it is of the form $p^{\prime} - P(z) p = Q(z)$ and you can use the integrating factor method shown on page 11 of this link. math.hawaii.edu/~jamal/tuc01alt_desupps.pdf $\endgroup$
    – mark leeds
    Aug 8 '18 at 4:49

The differential equation is of the form

$$y' + f(x)y = q(x)$$

The correct answer in our case is

$$p = -s$$

so that you know what you are targeting. Namely, it does not depend on $z$. You can verify that it satisfies the differential equation.

Then the author just plays around like

$$p= - s \implies p -p = s-s \implies p - \frac p z z = \frac s z z - s $$

$$ p = \frac p z z + \frac s z z - s = \frac {p+s}{z}z - s$$

$$\implies p = vz -s $$

...and I suppose this playing around with identities is helpful to what he does in the paper.

  • $\begingroup$ Hi Alecos: Two questions at your convenience of course. 1) If you wouldn't mind, could you define your $y^{\prime}$, $p(x)$ and $q(x)$ terms so that I can understand how mine was wrong. 2) Where did you get $p = -s$ from ? thanks. $\endgroup$
    – mark leeds
    Aug 9 '18 at 14:17
  • $\begingroup$ @markleeds There's nothing wrong with what you wrote. I wrote exactly the same thing that you did. Solving this differebtial equation, we get as solution $p=-s$. For the solution , see for example, mathworld.wolfram.com/… $\endgroup$ Aug 9 '18 at 14:19
  • $\begingroup$ Thanks. I'll check it out. Solving it looked hazardous so I didn't try !!!!! $\endgroup$
    – mark leeds
    Aug 9 '18 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.