How can the second order non-linear ODE be simplified using a substitution?

[talolard]

Homework Statement

yy''=y'^{2}-y'^{3}

I'm quite sure I got lost somewhere. Can anyone show me where?
Thanks

Set

z(y)=y'

then

\frac{\partial z}{\partial y}=y''\cdot y'=zy'' so y''=\frac{z'}{z}

Plugging this in

y\frac{z'}{z}=z^{2}\left(1-z\right) and so \frac{1}{y}\partial y=\frac{\partial z}{z^{3}\left(1-z\right)}

Integrating we have

ln\left(y\right)=\int\frac{1}{z^{3}}+\frac{1}{z^{2}}+\frac{1}{z}-\frac{1}{z-1}=-\frac{1}{z^{2}}-\frac{1}{z}+ln\left(\frac{z}{z-1}\right)+c

So

y=c_{1}\left(\frac{z}{z-1}\right)e^{-\left(\frac{z+1}{^{z^{2}}}\right)} and recalling that z=y' we have

y=c_{1}\left(\frac{y'}{y'-1}\right)e^{-\left(\frac{y'+1}{y'^{2}}\right)}

What now?

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

hi talolard!

\frac{\partial z}{\partial y}=y''\cdot y'=zy'' so y''=\frac{z'}{z}

no …

dz/dy = dz/dt dt/dy = y'' / y'​

(same as a = dv/dt = v dv/dx )

 

[talolard]

I wish I had asked sooner.
Thanks
Tal

 

[talolard]

yy''=y'^{2}-y'^{3}

Solution

Set

z(y)=y'

then

\frac{\partial z}{\partial y}=\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial y}=y''\cdot\frac{1}{y'}=y''\frac{1}{z}\rightarrow z\cdot z'=y''

Plugging this in and assuming z\neq0,1

yz\cdot z'=z^{2}\left(1-z\right)\iff yz'=z\left(1-z\right)\iff\frac{\partial z}{z\left(1-z\right)}=\frac{\partial y}{y}

Integrating we have

ln\left(y\right)=ln\left(\frac{z}{1-z}\right)+c\iff y=c_{1}\left(\frac{z}{1-z}\right)=c_{1}\left(\frac{y'}{1-y'}\right)

And here it gets tricky because i don't see a nice way to solve this. The best way I found is

\left(1-y'\right)=c_{1}\frac{y'}{y}=c_{1}\left(\ln\left(y\right)\right)'\iff c_{1}\left(lny\right)=x-y+c_{2}

Exponentiating we have

c_{1}y=c_{2}e^{x}e^{-y}\iff e^{y}y=\frac{c_{1}}{c_{2}}e^{x} where ignored the change in the constant.