When you say "the derivative of 0 is just 0", you are no longer in any way talking about the derivative of f(x) (as given in your problem). The derivative at a point depends on what your function is doing very close to that point as well. By "calculating" the derivative of f at zero the way...
The OP likely has some definition of what the expression \lim_{x\rightarrow\overline x}f(x)=L means. This question is merely asking them to show that whatever definition they have is equivalent to having f(x_n) converge to L for any sequence x_n\rightarrow\overline x.
We (those of us...
If your constant a_0 is positive, then indeed, those improper integrals converge. What you want is a technique referred to as Integration by Parts. This will allow you to find antiderivatives of functions of the form f(r)=r^n e^{-cr} among others, where c is a positive constant (so in your...
Having just looked up the history of his name, I now agree with you. I was under the impression that the English spelling L'Hospital came from the direction translation of Hôpital. Apparently he was named L'Hospital originally, with the silent s, but then the French changed their orthography...
Ah, true. Thanks for the clarification. And I really don't like including the s in L'Hôpital; personally I think dropping the accent and adding an s just makes the pronunciation even more confusing to newcomers. Don't want people to think he was a medical doctor or something :-p
Actually, it is L'Hopital's rule. Note that the limit might only exist if i and N have the same sign, or i is negative and N is positive (so as to guarantee we actually have an indeterminate form). We then have three cases; i<N, i=N, i>N.
Assuming the above limit exists (that is we're in the...
Your derivative of y_1 is incorrect. Your integrand is a function of x (and t), and your limits of integration are functions of x. See the "differentiation under the integral sign" formula: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign
For example, I find that...
I agree, as is, this isn't terribly convincing, unless Strogatz is talking about the global unstable manifold. Is this perhaps one of those rare examples where you can explicitly calculate the global unstable manifold? Sometimes this happens; the iteration method for finding the unstable...
I do not see how this equation is second-order. Where is the second derivative of y?
So where are you stuck? What are you doing?
It bothers me a little that you seem to be using an integrating factor on a nonlinear differential equation; typically, multiplying by an integrating factor is...
That's exactly what I got in the end. I plugged that function into Maple and checked the partial derivatives; it is definitely the solution. One final thing of course: since your solution doesn't depend on t, you can drop the dependence on it and safely write...
Yes, you're on the right track. Now, you have obtained u(t)=x_0e^{2x_0^2t}. This solution is not explicitly a function of x and y; it is a function of the parameter x_0 and the variable t. In the end, we want to get rid of the t's and x_0's and to somehow re-introduce the variables x,y.
We...
Without going into a full explanation on why the method works, what you typically do is: if your initial condition is written in the form u(x,x)=g(x), then your initial conditions for your characteristics become x(0)=x_0\\
y(0)=x_0\\
u(0)=g(x_0). Sometimes the initial condition might look...
To start, you haven't used the initial condition. Remember when you solve the characteristic equations (eg. the dx/dt=-x), these need initial conditions. In fact, the way you've written out your solutions for x and y, you seem to be assuming x(0)=1=y(0), which is not what you have.
Do you...
They're essentially equivalent, at least insofar as Kuratowski definition of ordered pairs is concerned. In this definition, the cartesian product of two sets X and Y, denoted by X\times Y is defined by the set of all ordered pairs (x,y)=\{\{x\},\{x,y\}\}, where the object on the left is just a...