Nonlinear 2nd order ode reduction solutions

[physicsjock]

hey guys

i've been trying to work out this ode reduction question,

http://img204.imageshack.us/img204/8198/asdawt.jpg

after i use the hint and end up with a seperable equation then integrate to get

[itex]\begin{align}
& p=\pm \frac{1}{\sqrt{{{y}^{2}}-2c}} \\
& \text{Then}\,\,\,\text{integrating}\,\,\,\text{again (using}\,\,wolframalpha) \\
& y=\pm \log \left( \sqrt{{{y}^{2}}-2c}+y \right)+c \\
& A{{e}^{\pm y}}=\sqrt{{{y}^{2}}-2c}+y \\
\end{align}[/itex]

the first line above is consistent with what i get when i use mathematica

but after getting p, is it correct to integrate p to get y? It doesn't look correct in my working but I can't find another way to do it.

i thought about trying to solve the equation as y'= +/- ... instead as a first order ode but it seems too complicated,

Is there something simple I am not seeing?

Thanks in advance

 

[HallsofIvy]

Yes, you are told that p= y'= dy/dx so
[tex]\frac{dy}{dx}= \pm\frac{1}{\sqrt{y^2- 2c}}[/tex]
and then
[tex]\frac{dy}{\pm\sqrt{y^2- 2c}}= dx[/tex]

Integrate both sides of that.

 

[physicsjock]

Hey HallsofIvy

Thanks for replying,

Are you sure its not:
[tex] \pm dy \sqrt{y^2- 2c}= dx[/tex]

since you take the dx to the right side, and the inverted root onto the left?

using the initial condition I get c=0 since

[tex]\begin{align}
& y'(0)=\pm \frac{1}{\sqrt{{{y}^{2}}(0)-2c}}=\pm \frac{1}{\sqrt{1-2c}}=-1 \\
& \pm \frac{1}{\sqrt{1-2c}}=-1,\,\,\,\pm 1=-\sqrt{1-2c}\,\,squaring\,\,both\,\,sides \\
& 1=1-2c \\
& c=0 \\
& So\,y'(x)=\pm \frac{1}{y} \\
& \pm ydy=dx \\
& \pm \frac{1}{2}{{y}^{2}}=x+c \\
& \pm \frac{1}{2}{{y}^{2}}(0)=\pm \frac{1}{2}=c \\
& \therefore y(x)=\sqrt{2x\pm \frac{1}{2}} \\
\end{align}
[/tex]

does that look ok?

 

[physicsjock]

I tried it again today with a fresh start and I ended up getting the same thing,
The thing that makes me sus about it is the +/- inside the square root