# Second derivative of an autonomous ODE

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.

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

AlephZero
Homework Helper
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
Homework Helper
Why not just seperate variables?

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).

Last edited:
AlephZero