Second order ODE reduction to first order solutions

hey,

I have this question I have been trying to solve, I got some help and I think I may have solved it but I just wanted to double check here ,

The question is, Consider the non-linear equation below, use y'=p to reduce it to first order, y(o)=1 and y'(0)=-1

y''+y(y')^{3}=0

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

so

$\displaystyle \pm dy \sqrt{y^2- 2c}= dx$

Then

$\displaystyle y'(0)=\pm \frac{1}{\sqrt{{{y}^{2}}(0)-2c}}=\pm \frac{1}{\sqrt{1-2c}}=-1$

$\displaystyle \pm \frac{1}{\sqrt{1-2c}}=-1,\,\,\,\pm 1=-\sqrt{1-2c}\,\,squaring\,\,both\,\,sides$

1=1-2c

c=0

$\displaystyle So\,y'(x)=\pm \frac{1}{y}$

$\displaystyle \pm ydy=dx$

$\displaystyle \pm \frac{1}{2}{{y}^{2}}=x+c$

$\displaystyle \pm \frac{1}{2}{{y}^{2}}(0)=\pm \frac{1}{2}=c$

$\displaystyle \therefore y(x)=\sqrt{2x\pm \frac{1}{2}}$

Does this look correct?

Thanks in advance

Re: Second order ODE reduction to first order solutions

Let $\displaystyle y'=v$ , where $\displaystyle v$ is a function with $\displaystyle y$ as argument ,then :

$\displaystyle y''=\frac{dv}{dy} \cdot v=v'_y \cdot v$

Hence :

$\displaystyle y''+(y')^3=0$

$\displaystyle v'_y \cdot v+v^3=0$

$\displaystyle v'_y=-v^2$

$\displaystyle \int \frac{dv}{v^2}=-\int dy$

Re: Second order ODE reduction to first order solutions

Isn't that the same result?

you would get

-1/v = -y

so y'= 1/y

only difference is i got +/-, is that the problem in my work having the +/-?

when I did it I had

y'' = dp/dy instead of (dp/dy)p

Re: Second order ODE reduction to first order solutions

I guess that your "$\displaystyle \pm $" isn't correct...

Re: Second order ODE reduction to first order solutions

Ahhh I left out a y!

What a silly mistake

The ODE should be

y''+y(y')3=0

Sorry,

$\displaystyle p'+yp{{'}^{3}}=0$

$\displaystyle \frac{p'}{{{p}^{3}}}=-y$

$\displaystyle \frac{dp}{{{p}^{3}}}=-ydy$

$\displaystyle -\frac{1}{2}\frac{1}{{{p}^{2}}}=-\frac{1}{2}{{y}^{2}}+c $

$\displaystyle {{p}^{2}}=\frac{1}{{{y}^{2}}-2c}$

$\displaystyle p=\pm \sqrt{\frac{1}{{{y}^{2}}-2c}}$

Thanks how I got the +/-

Re: Second order ODE reduction to first order solutions

I suggest you to use method that I used..It is a standard method for ODEs of this type .

Re: Second order ODE reduction to first order solutions

So using your method the solution would be

$\displaystyle y(x)=\sqrt{2x + \frac{1}{2}}$

Is that right?

Edit:

Hold on that's not right, just working it out quickly now

I get

y^{3 }- 3/2 y = 6x - 1/2

Does that look right?

Re: Second order ODE reduction to first order solutions

Re: Second order ODE reduction to first order solutions

...did not know it could do that haha,

so its y^{3} +5/2 y = 6x + 3/2

thanks!

Re: Second order ODE reduction to first order solutions

You could give a name to this thread "Thanksgiving thread "....(Giggle)