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 Lipschitzcondition. 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?

Maple, though, finds an implicit solution. Plus, this is just a separable equation, which should be straightforward to solve. – Artem Apr 27 '12 at 15:10

It's usually a good idea when referring to WolframAlpha results to link to them. In the present case, it seems you entered the equation either incorrectly or in an inconvenient form; when I enter it, I get [an implicit solution](http://www.wolframalpha.com/input/?i=solve+y%27%3Da+y^2+%2F+2+%2B+y^%283%2F2%29). – joriki Apr 27 '12 at 15:51

Ouch, that is actually easy to see. I was so puzzled by this $W$function. However, this was my input: [WAInput](http://www.wolframalpha.com/input/?i=DSolve%5By%27%5Bx%5D+%3D%3D+1%2F2*a*y%5Bx%5D%5E2%2B%28y%5Bx%5D%29%5E%283%2F2%29%2C+y%5Bx%5D%2C+x%5D). To my knowledge, this is standard mathematica input. – user13655 Apr 27 '12 at 16:22

The 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 $y$. – Robert Israel Apr 27 '12 at 16:26
1 Answers
The standard way to solve such a separable firstorder ordinary differential equation is to divide through by the function of $y$ to get separate integrals with respect to $t$ and $y$:
$$ \frac{y'}{\frac12\alpha y^2+y^{3/2}}=1\;, \\ \int\frac{y'}{\frac12\alpha y^2+y^{3/2}}\mathrm dx=\int1\,\mathrm dt\;, \\ \int\frac1{\frac12\alpha y^2+y^{3/2}}\mathrm dy=\int1\,\mathrm dt\;. $$
WolframAlpha 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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@user13655 I think joriki has other standards for [extensive](http://math.stackexchange.com/questions/33094/deletinganydigityieldsaprimeisthereanameforthis/33303#33303) – Pedro Apr 28 '12 at 16:35