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

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

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
$$\frac{dy}{dx}= \pm\frac{1}{\sqrt{y^2- 2c}}$$
and then
$$\frac{dy}{\pm\sqrt{y^2- 2c}}= dx$$

Integrate both sides of that.

 

[physicsjock]

Hey HallsofIvy

Thanks for replying,

Are you sure its not:
$$\pm dy \sqrt{y^2- 2c}= dx$$

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

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

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