Hello, I am trying to solve the following equation:
\frac{\partial x}{\partial t} = A \frac{ \partial x}{\partial y} + B \frac{\partial^2 x}{\partial y^2}
I know how to solve the diffusion equation (i.e. no dx/dy term), but that method doesn't work here. I tried to go with the LaPlace...
I haven't done this in ages, and I'm having trouble recalling how to factor a higher order polynomial. I almost always do this graphically, but for this case I'm interested in an algebraic solution. Specifically, I'm looking at ax + x^3 - x^5 = 0 (with a = an integer >0.)
Clearly 0 is one...
I think I have it - this means that:
\frac {1}{v} \frac {1}{v+b} = \frac {A(v+b)+Bv}{v(v+b)}
This would imply, since the denominators are the same, that:
1 = A(v+b)+Bv.
I can then solve for A and B by choosing easy values for v. So, if v=0, then A=1/b. If v=-b, then B=-1/b. Plugging that...
In my continued pursuit of isothermal equation of state solutions, I've come upon a very simple integral I can't recall how to do, and was looking for assistance.
It is:
du = \int \frac{a}{T^{0.5}v(v+b)}dv
where a, T and b are constants. so:
du = \frac{a}{T^{0.5}} \int \frac{1}{v(v+b)}dv...
Ah, I see. Thanks for the help!
(of course, since you are familiar with the physics, Gib z, you may see I'm trying use the relation of your namesake to fund du, and I could do the attraction potion which is the ultimately simple a/v^2.
Again, I greatly appreciate the help!
I'm trying to integrate the Van der Waals equation of state for an isothermal problem, but based on my results I think I'm doing a bit of simple calculus wrong and hope someone here can help.
P = \int \frac{RT}{v-b} dv
where R, b and T are known constants.
I tried to do a u-substitution...
Simple Diff Eq Help
I am trying to solve the following Diff Eq:
\frac{d^2x}{dy^2}+(\frac{y}{2}-\frac{1}{y})\frac{dx}{dy}=0
I tried to solve by setting \frac{dx}{dy}=z
so: \frac{dz}{dy}+(\frac{y}{2}-\frac{1}{y})z=0
I know the general solution to this is...
Okay, so looking at this backwards makes sense.
Since w is general, and thus can be f(x,y,z,t)
\frac {D \frac{\omega^2}{2}}{Dt} differentiated by the chain rule gives:
\frac{2 \omega}{2} \frac{D \omega}{Dt}
= \omega \cdot \frac{D \omega}{Dt}
So, looking at it backwards makes it...
What I would try is:
1. Come up with a time dependant expression of the volume of water in the bottle, based on the starting volume of water minus the volumetric flow rate of water leaving the bottle.
2. Then, find an expression for the hydrostatic pressure the water is exerting on the...
Try being nice to her. But talk about your girlfriend anytime you see her. She'll get the hint.
For some reason I don't comprehend, many girls often enjoy the chase for attention, and the harder it is to get, the harder they pursue it (much as Twisting Edge says.) But, the same sort hate...
1. Capital D refers to the substantial derivative, in the notation of Stokes. It boils down to:
\frac {D()}{Dt} = \frac {\partial ()}{\partial t} + v_i \partial_i ()
Also, in the index notation of my book, \frac {\partial ()}{\partial t} = \partial_o
2. w is a vector. v is a...
In my struggles to understand index notation, I am trying to figure out how my book came up with the following transformation.
\frac {D \omega}{Dt} \cdot \omega = \omega_j \partial_j v_i \cdot \omega + \nu \partial_j \partial_j \omega_i \cdot \omega
=
\frac {D \frac{\omega^2}{2}}{Dt} =...