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PostPosted: Mon, 29 Mar 2010 19:51:00 UTC 
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I was looking at Exam A from the Putnam from last December and one of the problems looked like a fun thing to spend a couple of minutes trying. It's not hard by any stretch of the imagination, but the method was rather novel, and I think it will be fun for those with even the most rudimentary understanding of DiffEQ.

Functions f,\, g, and h are differentiable on some open interval around 0 and satisfy the equations and initial conditions

\\
f'=2f^2gh+{1\over gh}\quad f(0)=1 \\
g'=fg^2h+{4\over fh}\quad g(0)=1  \\
h'=3fgh^2 + {1\over fg}\quad h(0)=1

Find an explicit formula for f(x) valid on some open interval around 0.

Hint 1:

Spoiler:
Product rule


Hint 2:
Spoiler:
semi-clever replacement


My solution:

Spoiler:
Start by noting that the identically zero function for any of the three is invalid by the division part of the equations, so multiplying all of the equations by f,\, g, or his a valid operation.

Then we have:


\\
f'gh=2(fgh)^2+1\quad f(0)=1 \\
fg'h=(fgh)^2+4\quad g(0)=1  \\
fgh'=3(fgh)^2 + 1\quad h(0)=1

Add these together and note that we are in a situation where we have applied the product rule to the function m:=fgh, so we get:

m'=6(m-1)

The semi-clever replacement is to note that Dm=D(m-1)

So solving the differential equation:

D(m-1)-6(m-1)=0 is simple, it gives that

m=Ae^{6x}-1 by use of the characteristic equation method.

Now the initial condition m(0)=1 tells us that A=2.

Now returning to the equation

${df\over dx}=f\left(2m+{1\over m}\right)\quad f(0)=1

we can see by simple separation of variables that

$\log f=2\int (2e^{6x}-1)\,dx +\int{dx\over 2e^{6x} -1}

which yields

$\log f = \left({2\over 3}e^{6x}-2x\right)+\left({1\over 6}\log{\left| 1-2e^{6x}\right|}-x\right)+C

The second integral coming from adding and subtracting 1 and making a u-substitution.

Using our initial conditions, we find that 0={2\over 3}+C\Rightarrow C=-{2\over 3}

From there it's a simple exponentiation to recover f completely.

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