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I have a question regarding an example problem at about 22min on this lecture http://ocw.mit.edu/courses/mathemat...ecture-2-eulers-numerical-method-for-y-f-x-y/

The equation in question is y'=x[itex]^{2}[/itex]-y[itex]^{2}[/itex].

In an example regarding the Euler method, Prof. Mattuck describes the second derivative of the above function as

y'' = 2x - 2yy'

specifically mentioning the chain rule. Now, as I understand it (working backwards of course), the only way to this solution would be to consider y[itex]^{2}[/itex] as the function y(y(x)), giving the solution

yy' + y'y = 2yy'.

Now, assuming I correctly understand how the 2yy' portion of the solution was derived, my question is: how exactly was the second derivative determined to be

y""= 2x - 2yy' ?

If we take the derivative with respect to y (which presumably give us the 2yy' term we're looking for), wouldn't the "x" term (x[itex]^{2}[/itex]) be zero, as shown:

d/dy (y')= d/dy (x[itex]^{2}[/itex])- d/dy (y[itex]^{2}[/itex]) [itex]\Rightarrow[/itex] y''= 0 - 2yy'

Wolfram Alpha also seems to agree with me, but I'm afraid I'm more muddled than I'd like to believe

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# Second derivative - wolfram giving one answer, professor giving another

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