Data said:
Yeah, the resulting integral for y is not nice at all. But at least he has it down to an integral, which is perfectly sufficient for approximating a solution :).
Thanks Data. When I first saw the problem, I didn't think to solve it that way. Just in case you guys don't feel like typing it all in, I'll do the next part. Perhaps Maphysique will follow it through:
I'll write the exact equation as follows:
[e^{4y}(2v^2 + y]dy + [e^{4y}v]dv = 0
"Exact" means its a total differential of some function F(y,v), that is:
dF=[e^{4y}(2v^2 + y]dy + [e^{4y}v]dv
Then surely:
\frac{\partial F}{\partial y}=[e^{4y}(2v^2 + y]
and:
\frac{\partial F}{\partial v}=[e^{4y}v]
To find F(y,v), let's integrate the partial with respect to y:
\int{\frac{\partial F}{\partial y}=\int{[e^{4y}(2v^2 + y]}dy
And thus:
F(y,v)=\frac{1}{2}v^2e^{4y}+\frac{1}{4}ye^{4y}-\frac{1}{16}e^{4y}+T(v)
Where integrating the partial with respect to y necessarilly yields some arbitrary function of v, T(v).
Differentiating F(y,v) with respect to v now yields:
\frac{\partial F}{\partial v}=ve^{4y}+\frac{dT}{dv}
But according to the exact differential, this is also equal to e^{4y}v
Or:
ve^{4y}+\frac{dT}{dv}=e^{4y}v
or:
\frac{dT}{dv}=0
Integrating yields:
T(v)=k
So that:
F(y,v)=\frac{1}{2}v^2e^{4y}+\frac{1}{4}ye^{4y}-\frac{1}{16}e^{4y}+k
Where k is a constant which we'll absorb into the constant c:
\frac{1}{2}v^2e^{4y}+\frac{1}{4}ye^{4y}-\frac{1}{16}e^{4y}=c
Finally, solving for v (which is y') yields:
(\frac{dy}{dx})^2=ce^{-4y}-\frac{1}{2}y+\frac{1}{8}
or:
\frac{dy}{dx}=\pm \sqrt{ce^{-4y}-\frac{1}{2}y+\frac{1}{8}}
But that's ok. You just solve two differential equations now and use whatever part meets the initial conditions. Anyway, I'll spend some time studying this numerically in Mathematica and report some useful results with select initial conditions.
