# Nonlinear 2nd order ode reduction solutions

1. Mar 30, 2012

### physicsjock

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

\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}

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?

Last edited by a moderator: May 5, 2017
2. Mar 30, 2012

### HallsofIvy

Staff Emeritus
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.

3. Mar 30, 2012

### physicsjock

Hey HallsofIvy

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} & 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}

does that look ok?

Last edited: Mar 30, 2012
4. Mar 31, 2012

### 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