finding a singular solution

khauna · #1 $Old$ November 2nd, 2009, 11:16 PM

I am having trouble with this problem. I believe that there is a singular solution involved in it yet I do not know how to find it.

(z'')(z^2)=(z')^3

so far i have rewritten the problem as
(z'')(z^2)-(z')^3=0 , made the substitution u=z' and continued to solve that using separation of variables. I still do not understand how to find the singular solution though

thank you for any help

shawsend · #2 $Old$ November 3rd, 2009, 07:26 AM

How about we write it as just $y'' y^2=(y')^3$

By inspection you can see $y=k$ is a solution including the solution $y=0$ right? So letting $y'=p$ and making that substitution, I get $pp'y^2=p^3$ and I can do the division to obtain $\frac{dp}{dy}p^{-2}=y^{-2}$ only if $p\neq 0$ and $y\neq 0$ . Doing that, and integrating twice, I obtain the solution:

$c_1 y+\ln(y)=x+c_2$ . Now note that the solutions $y(x)=k$ are not particular cases of this solution and it looks like the general solutions are asymptotically tangent to the solution $y=0$ but not the other y=k solutions. I'm a little unsure about this part as to whether all the y=k solutions are singular or just the y=0 one.