Second derivative of an autonomous ODE

[MurdocJensen]

Alright. So I have dy/dx = -1-y². I want to take the second derivative to get some information about the concavity of the solution, but I can't wrap my head around what's really going on.

What I think I know: I have an ODE that is dependent on the dependent variable, so my solution will only change with changes in the value of y.

What I want to know: How is dy/dx (the derivative of y(x) with respect to x) dependent completely on y? How do I take the second derivative with respect to x when I have no x in the first derivative? Do I have an x nested in y in the derivative?

I get the feeling I'm missing something really simple. If you find it enough to just give me a hint, I'm definitely cool with that. More in depth answers are awesome too.

 

[l'Hôpital]

Remember that y still depends on x. So, use the chain rule.

 

[AlephZero]

After you have differentiated the equation (using the chain rule), you can then use the original equation to eliminate dy/dx.

That will give you d^2y/dx^2 = a function of y only.

 

[lurflurf]

Why not just separate variables?

 

[MurdocJensen]

Thanks for the information. Here's what I have:

1) dy/dx = -1-y²

2) d²y/dx² = -2y(y') = -2y(-1-y²) = 2y+y³

Yea?

So I am technically differentiating a derivative with respect to x. Because this D.E. is autonomous, doing this means I need to differentiate, using the chain rule, with respect to y (a function whose independent variable is x).

 

[AlephZero]

That looks good to me.