<p>The differential equation is of the form</p>

<p>$$y' + f(x)y = q(x)$$</p>

<p>The correct answer in our case is</p>

<p>$$p = -s$$</p>

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

<p>Then the author just plays around like</p>

<p>$$p= - s \implies p -p = s-s \implies p - \frac p z z = \frac s z z - s $$   </p>

<p>$$           p = \frac p z z + \frac s z z - s = \frac {p+s}{z}z - s$$</p>

<p>$$\implies p = vz -s $$</p>

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

CC BY-SA 4.0
2018-08-09T14:17:08.060Z
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.
mark leeds
CC BY-SA 4.0
2018-08-09T14:19:53.330Z
@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, http://mathworld.wolfram.com/First-OrderOrdinaryDifferentialEquation.html
Alecos Papadopoulos
CC BY-SA 4.0
2018-08-09T14:22:03.087Z
Thanks. I'll check it out. Solving it looked hazardous so I didn't try !!!!!
mark leeds
CC BY-SA 4.0
2018-08-09T11:28:22.530Z
2018-08-09T14:18:46.433Z
2018-08-09T14:18:46.433Z
Alecos Papadopoulos
<p>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: 
$$\frac{dp}{dz} = (p+s)/z.$$</p>

<p>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? </p>

CC BY-SA 4.0
2018-08-08T04:49:47.197Z
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. http://www.math.hawaii.edu/~jamal/tuc01alt_desupps.pdf
mark leeds
<microeconomics><difference-equations>
Solving differential equation in The Economics of Superstar (by Rosen)
shk910
# Solving differential equation in The Economics of Superstar (by Rosen)

(Asked by: username_0 on 2018-08-08)

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?

**Tags:** microeconomics, difference-equations

> **Comments:**
> **username_1**: 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. http://www.math.hawaii.edu/~jamal/tuc01alt_desupps.pdf  

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## 1 answers

### Answer by username_2 on 2018-08-09. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_1**: Hi username_2: 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.  
> **username_2**: @username_1 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, http://mathworld.wolfram.com/First-OrderOrdinaryDifferentialEquation.html  
> **username_1**: Thanks. I'll check it out. Solving it looked hazardous so I didn't try !!!!!  

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