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hey guys

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

http://img204.imageshack.us/img204/8198/asdawt.jpg [Broken]

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 im not seeing?

Thanks in advance

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

http://img204.imageshack.us/img204/8198/asdawt.jpg [Broken]

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 im not seeing?

Thanks in advance

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