In a paper the differential equation $-y'(t)=\frac{1}{2}\alpha y(t)^2+y(t)^{3/2}$ occured. First of all, it does not satisfy the usual Lipschitz-condition. But there are probably weaker criteria. The solution was given as: $ y(t)=\frac{4}{\alpha^2}\left(1+W(-C_1 e^{-\frac{t}{\alpha}})\right)^{-2}, $ where $W$ is the Lamber-$W$-function. Mathemica/WolframAlpha give nothing. I have not yet seen such a differential equation. How can one solve this?
Nonlinear differential equation: $-y'(t)=\frac{1}{2}\alpha y(t)^2+y(t)^{3/2}$
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0The right side of the equation is locally Lipschitz (on $[0,\infty)$). To avoid problems with negative $y$, replace the $y^{3/2}$ on the right side with $|y^{3/2}|$. Since this modified equation satisfies the hypotheses of the existence and uniqueness theorem, and $y=0$ is a solution, the solution with any initial value > 0 will never hit $0$, so you don't have to worry about negative $y$. – 2012-04-27
1 Answers
The standard way to solve such a separable first-order ordinary differential equation is to divide through by the function of $y$ to get separate integrals with respect to $t$ and $y$:
$ -\frac{y'}{\frac12\alpha y^2+y^{3/2}}=1\;, \\ \int-\frac{y'}{\frac12\alpha y^2+y^{3/2}}\mathrm dx=\int1\,\mathrm dt\;, \\ \int-\frac1{\frac12\alpha y^2+y^{3/2}}\mathrm dy=\int1\,\mathrm dt\;. $
Wolfram|Alpha solves the integral on the left (and shows steps if you click "Show Steps"):
$\frac2{\sqrt y}-\alpha\log\left(\alpha+\frac2{\sqrt y}\right)=t+C\;.$
With $u=\alpha+2/\sqrt y$, this becomes
$u-\alpha\log u=t+C'\;.$
Rescaling by $u=-\alpha s$ to get rid of the factor $-\alpha$, we get
$s+\log s=-\frac t\alpha+C''\;.$
Then exponentiating yields
$s\mathrm e^s=C'''\mathrm e^{-t/\alpha}\;,$
which is solved using the Lambert W function by
$s=W\left(C'''\mathrm e^{-t/\alpha}\right)\;.$
Then resolving the substitutions yields
$y=\left(\frac2{u-\alpha}\right)^2=\left(\frac2\alpha\right)^2\left(1+s\right)^{-2}=\frac4{\alpha^2}\left(1+W\left(C'''\mathrm e^{-t/\alpha}\right)\right)^{-2}\;.$
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0@user13655 I think joriki has other standards for [extensive](http://math.stackexchange.com/questions/33094/deleting-any-digit-yields-a-prime-is-there-a-name-for-this/33303#33303) – 2012-04-28