# Thread: finding a singular solution

1. ## finding a singular solution

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

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

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

$\displaystyle c_1 y+\ln(y)=x+c_2$. Now note that the solutions $\displaystyle y(x)=k$ are not particular cases of this solution and it looks like the general solutions are asymptotically tangent to the solution $\displaystyle y=0$ 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.