I am assigning an economics problem to my undergrads, and I have reduced the problem to a differential equation which I do not know how to solve. I can leave the problem for them at the point where the have to find the diff-eq but I would like to present the solution to the equation to them when we go over the problem set. Can anyone help?

Here is the equation:

Find p(v):

(1-v)(v-2p)-(vp-p^2)/p'=0
p(0)=0

The origin of the problem is in 2 sided bargaining. 2 players with valuations drawn from a uniform distribution make Take it or leave it offers to a seller whose valuation is also uniform on zero one.

Any help would be much appreciated.

Sorry, it's actually

v(v-2p)+(vp-p^2)/p'=0
p(0)=0

Well, this does make quite a difference. The latest equation is homogeneous. A substitution

p = vq(v)

leads to a separable equation for the function q. Which, in principle, can be solved by common integration. I have not checked that the resulting integral is simple or expressible in terms of elementary functions.

Also, at an academic institution, there must be available some kind of symbolic integration package: Mathematica, Matlab, or something more specialized. I would check.

I looked at this for a while yesterday, and made little progress, eventually concluding that this differential equation was not amenable to any of the tricks in my repertoire. The revised version is, as Alex says, homogeneous. I looked at the form you get if you make the substitution p = vq, and the resulting integrands do simplify in a nice way if you make the further substitution r = q-1/3, which comes from completing the square on the denominator after variable separation. The result is

(1-6r)dr/(r^2-1/9) = dv/v.

The left side integrates easily to one trig term and one reciprocal term. This will give an implicit function for p in terms of v, which would be rather hard (I think) to solve explicitly for p. On the other hand, you can perhaps get v explicitly as a function of p, although the substitution p=qv will make this awkward. In any case, the differential equation is "solved" in the formal sense that the expression relating p and v now contains no derivatives, although extracting the value of p for a given v will still require solving a rather messy algebraic equation. I suspect that plotting the result numerically will be the best approach.

Hope that helps!

Stuart Anderson

v(v-2p)+(vp-p^2)/p'=0
Since this is an homogeneous ODE and p(0)=0 then p=kv
v(v-2kv)+(v(kv)-(kv)^2)/k=0
(v^2)(2-3k)=0
k=2/3
The solution is : p=2v/3
.
For information, the complete resolution is:
v(v-2p)p'+(v-p)p=0
p=vt
p'=t+vt'
v(v-2vt)(t+vt')+(v-vt)vt=0
(1-2t)(t+vt')+(1-t)t=0
(1-2t)(t')+t(2-3t)/v=0
((2t-1)/(t(3t-2)))t'=-1/v
(1/2)ln(t)+(1/6)ln(3t-2)=-ln(v)+c
(3t-2)(t^3)=C/(v^6)
(3(p/v)-2)(p^3)/(v^3)=C/(v^6)
(3p-2v)(p^3)=C/(v^2)
3(p^4)-2v(p^3)-C/v^2=0
p=p(v) can be obtained by solving this quartic equation (unknown p) using the arduous Ferrari's formulae.
In particular case C=0, the equation reduces to (p^3)(3p-2v)=0 leading to the obvious particular solution p=2v/3.

@JJ

I followed the same process as you did, up until the point where you did partial fractions, while I completed the square on the denominator of the integrand. (Also, I forgot a factor of 9 in my integrand, which of course would not affect its integrability.) However, after that, the forms of the integral look quite different, although they must be equivalent. Yours appears to be nicer.

--Stuart Anderson