Solving 2nd Order ODE: yy``=2y`^2

[asdf1]

for the following question:
yy``=2y`^2

my problem:
i don't have a clue how to get a hand on this one! any suggestions?

 

[Benny]

yy'' = 2\left( {y'} \right)^2. The independent variable doesn't seem to be there. So perhaps p\left( y \right) = y' \Rightarrow y'' = p\frac{{dp}}{{dy}} so that yp\frac{{dp}}{{dy}} = 2p^2. It would also be a good idea to not 'cancel' a p from both sides.

 

[asdf1]

@@a
may i ask:
how'd you think of that?

 

[HallsofIvy]

That's a fairly standard "reduction of order" method.

If y"= f(y, y') so that there is no x explicitely in the equation, then
Letting u= y' gives, by the chain rule, y"= u'= (du/dy)(dy/dx)= u(du/dy) resulting in the first order equation u(du/dy)= f(y,u) with y as the independent variable and u as the dependent variable.

 

[saltydog]

yy'' = 2\left( {y'} \right)^2. The independent variable doesn't seem to be there. So perhaps p\left( y \right) = y' \Rightarrow y'' = p\frac{{dp}}{{dy}} so that yp\frac{{dp}}{{dy}} = 2p^2. It would also be a good idea to not 'cancel' a p from both sides.

You can divide by p if p is not zero. If p=0 then that means y=constant. Note that this solution satisfies the ODE.

 

[asdf1]

thank you!