OK, I think I answered my own question.
Do a binomial expansion on the polynomial term. The kth term in this expansion is given by:
(x-t)^n = \left(\array{c}n \\ k\endarray\right)x^{n-k}(-t)^k
so the function Y can be rewritten as the sum:
Y(x) = \frac{1}{n!}\int_0^x...
Is there a simple way to show that when we differentiate the following expression (call this equation 1):
Y(x) = \frac{1}{n!} \int_0^x (x-t)^n f(t)dt
that we will get the following expression (call this equation 2):
Y'(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t)dt
It's simple...
I'm reading Ince on ODEs, and I'm in the section (in Chapter 5) where he talks about the Wronskian. There are quite a few things here that I don't quite understand or follow.
I'm not going to get into all the details, but briefly, suppose we have the Wronskian of k functions:
W =...
Yes, it is interesting.
It's also interesting to consider the geometrical implications of each solution.
The first solution is as follows:
\sin^{-1}x + \sin^{-1}y = c_1
or in other words, if we define two angles u and v such that:
u = \sin^{-1}x
v = \sin^{-1}y
then this...
I'm not sure I completely understand the problem statement, but in general on a circle, the radius is normal to the curve.
So if you're interested in "normal" derivatives, you usually only have to consider how things change w/ respect to radius. This is something along the lines of the...
It's interesting to note, in passing, that if we take the "two" solutions for the D.E. given above, and calculate their Jacobian, the answer is 0, which is what we would expect if the "two" solutions were actually the same, and just differed in how they were represented.
Differential...
Yes, but the third equation is a solution for the second (and/or first), and indeed you've indicated the method for how to get from the second equation to the third, which was my original question: the "integrating" factor is just (much) more complex than I had thought.
You start with...
Suppose we have the differential equation:
\frac{dx}{\sqrt{1-x^2}} + \frac{dy}{\sqrt{1-y^2}} = 0
It can be rewritten as:
\sqrt{1-y^2}dx + \sqrt{1-x^2}dy = 0
One solution of this equation (besides arcsin(x) + arcsin(y) = C), is given by:
x\sqrt{1-y^2} + y\sqrt{1-x^2} = C...
In well-known case of the Clairaut equation, the singular solution forms an "envelope" or "boundary" around the family of general solutions to the D.E.
Is this always the case (i.e., do singular solutions always form a boundary around the set of general solutions), or does this only happen to...
Yes, I'm familiar w/ Wikipedia.
If there's a specific example from Wikipedia that you'd like to discuss in more depth, please do so.
Getting back to my example:
(x+y)^2y' = 0
The equation is solved by "two" equations:
y_1 = -x
y_2 = C
Suppose we set initial conditions to...
Just so I have the concept of a singular solution down correctly, suppose I have an equation like:
\left(x+y\right)^2y' = 0
This admits of two solutions:
y=-x
and, from:
y' = 0
y = C
where C is a constant.
So the "two" solutions for the equation would be:
y_1=-x, y_2 =...
I think the answer is "yes"..
For instance, the Euler theorem on homogeneous functions states, in relevant part, that if you have a function in two variables, x and y, which is homogeneous in degree n, then:
n f(x,y) = x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y}
So...
Is there such a thing as a homogenous function of degree n < 0?
Considering functions of two variables, the expression:
f(x,y) = \frac{y}{x}
is homogeneous in degree 0, since:
f(tx,ty) = \frac{ty}{tx} = \frac{y}{x} = f(x,y) = t^0 \cdot f(x,y)
and the expression:
f(x,y) = x...
You're correct about deriving the associated Legendre equation.
I'll step through the math (for my own edification, if nothing else, and for the sake of completion), and then try to answer your question in the next post.
First, starting with:
\Theta'' + \cot \theta \cdot \Theta' +...
True, but the reason I'm writing it all out is to make sure I've got all the equations right.. :)
Easy to get jumbled up in all this..
I'm not sure I'm getting the same expression you are for the \Theta expression.
Recapping where we are, we've gotten the expression separated out as...
Let's keep working w/ the second equation, Y, which, as I indicated earlier, can be separated much like the radial component was separated:
Y(\theta,\phi) = \Theta(\theta)\Phi(\phi)
Plugging this into the above expression for Y, we have:
\sin^2\theta \cdot \mu + \frac{\sin...
What you're trying to do is solve the equation:
\nabla^2u = -ku
where u is expressed in spherical coordinates, and where the Laplacian takes the form:
\nabla^2 u = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial u}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial...
Are you able to find solutions for the Legendre equation?
\left[(1-x)^2y'\right]' + \mu y = 0
I ask this, b/c you should be to separate the variables a second time; that is, to assume that:
Y(\theta,\phi) = \Theta(\theta) \Phi(\phi)
If you plug in this for your second above, you'll...
I'm reading a text on PDEs..
I'm trying to follow some of the argument the author is presenting, but I'm having a bit of difficulty.
We start w/ a collection of p functions in n variables (with p <= n). That is to say, we have:
u_1, u_2, ..., u_p
where
u_i : \mathbb{R}^n...