- #1
boredat20
- 10
- 0
Hi everybody.
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
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