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

so

Then

1=1-2c

c=0

Does this look correct?

Thanks in advance

Re: Second order ODE reduction to first order solutions

Let , where is a function with as argument ,then :

Hence :

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 " " 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,

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

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)