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    3 Problems involving superposition

    Isn't the \lambda for the third harmonic \frac{2L}{3}? If you use hookes law, don't we need to know the spring constant k before using this information to solve this problem?
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    Maxwell's Speed Distribution

    Sorry to keep badgering you here, but I do not see how you get \frac{1}{2}C^{-3/2}\int_{0}^{\infty}\sqrt{x} e^{-x}dx The \sqrt{x} should be an x/C no? We let x = Cv², so that makes v² = x/C. The term you are substituting for there is v² not v. That would make the integral...
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    Maxwell's Speed Distribution

    Oh ok, so you are saying that the original equation should be: \int 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{2RT}} dv = 1 instead of: \int 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{nRT}} dv = 1 ?
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    Maxwell's Speed Distribution

    Thanks for the reply, but I still seem to be a little lost. If I let C = \frac{M}{2RT} , that leaves me with 4 \pi \cdot ( \frac{C}{ \pi})^{3/2} \int \frac{x}{C} ... . I don't see how making the C = \frac{M}{2RT} helps me in the latter half of the integral because I have a n term in there...
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    Maxwell's Speed Distribution

    Homework Statement Given Maxwell's probability distribution function, P(v) = 4\pi\ (\frac{M}{2\pi RT})^{3/2} \cdot v^2 \cdot e^{\frac{-Mv^2}{nRT}} Where v = velocity, M = molar mass, R = Universal Gas Constant, n = # of mols, T = temperature, solve \int P(v) dv =1 from 0 to...
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    Experimental Proof of Electrical Charge

    Hi, I have a question about the transfer of electrical charge from object to another. Basically, my professor stated that if you rub a rod with certain matierials, the rod will become charged. This is due to the convention that Ben Franklin came up with called the triboelectric series. So...
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    Find the Potential as a function of position

    Homework Statement A rod of length L carries a charge Q uniformly distributed along its length. The rod lies along the y-axis with one end at the origin. Find the potential as a function of position along the x-axis Homework Equations dV=\vec{E}\cdotp d\vec{l} V=\frac{kq}{r}...
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    Explaination of Curvature

    Awesome, thanks guys. You really cleared this up for me!
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    Explaination of Curvature

    Homework Statement Find the curvature of y = x³ Homework Equations k(x) = \frac{f"(x)}{[1+(f'(x))²]^{3/2} The Attempt at a Solution k(x) = \frac{6x}{(1+9x^4)^{3/2} I got the answer numerically, but I am looking for an explaination of the graph itself. I chose a relatively...
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    General initial value problem (DE's)

    Homework Statement a) Consider the initial value problem \frac{dA}{dt} = kA, A(0) = A_0 as the model for the decay of a radioactive substance. Show that in general the half-life T of the substance is T = -\frac{ln2}{k} b) Show that the solution of the initial-value problem in part a) can...
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    Help with a simple Diff Eqn

    Oh cool, even better. Thanks!
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    Ok, so now how do you separate this one?

    Yes you did AKG, sorry about that. I just did not see how to factor them. I guess I am not very strong with my factoring and it was not very clear to me. Thanks for the hint Daniel, that will help me out a lot. Ok, going to go and work this out now!
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    DE with an initial condition

    I am asked to solve this DE with the initial condition of y(1) = 1. (x+y)^2dx + (2xy + x^2-1)dy = 0 So, after working the problem out, I came to this as an answer: F(x,y)=\frac{1}{3}x^3 + x^2y + xy^2-y My question is what do I do with the initial condition. I assume that I am just...
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    Explanation of this Partial?

    \frac{\partial_P}{\partial_y}(2ysinxcosx-y+2y^2e^{(xy^2)} I worked the first part no problem, but the second part I needed a little help from my calculator. This is what I got: 2sinxcosx-1+4ye^{(xy^2)} My question is, why does the partial of 2y^2e^{(xy^2)} come out to 4ye^{(xy^2)}...
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    Ok, so now how do you separate this one?

    ok, so I tried it and I got 1+\frac{5(x-y+1)}{xy-2x+4y-8} which does not seem to help me out too much. The form of both the numerator and the denominator do look a little suspicious. Is there a way to factor them like we can for a problem in the form ax^2 + bx + c?
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    Ok, so now how do you separate this one?

    Could you please be a little more specific? Do you mean just divide the two using long division? Hmmm... ok, I will try that and see if I can get to something. Thanks.
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    Did I solve this DE correctly?

    Also, one quick question. Does the equation have to be in the form of P(x.y)dx + Q(x,y)dy = 0? Can it be minus instead of plus? The reason I ask is because I vaugly remember hearing something about that at the begining of the school quarter and can't seem to find it in my notes now. Thanks!
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    Did I solve this DE correctly?

    If someone has a chance out there, could you please check my math here and let me know if I am doing this correctly or not. Problem: Solve: (2x-1)dx + (3y+7)dy = 0 I would like to solve this using the "Exact" method for solving DE's, so: \frac{\partial_P}{\partial_y}(2x-1) = 0...
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    Ok, so now how do you separate this one?

    umm... Sorry if this sounds lame, but I don't see what I am to factor here.
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    Ok, so now how do you separate this one?

    \frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8} I want to solve this DE using the separation technique. Any ideas on how to start? And just for myself, and maybe anyone else. Is there a sort of systematic approach to finding out how to start these problems? It seems like I will work a few no...
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    Idea of how to separate?

    Oh man, haha.. I like the substitution method, but I am not too keen on the trig substitution. Although, this one seems pretty nice for that. Do you mind taking a look at my partial fraction work? I need some practice with it so I was wondering if I did it correctly...
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    Idea of how to separate?

    Thanks, I will try that out. To get the LHS w/o partial fractions: \int{\frac{ydy}{(y+1)^2}} let u = y+1, du = dy, and y = u-1 So, \int{\frac{(u-1)}{(u)^2}du} = \int{\frac{u}{u^2}-\frac{1}{u^2}du} = ln|u| + \frac{1}{u} Substitute y+1 for u and get ln|y+1|+\frac{1}{y+1}
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    Idea of how to separate?

    Yes I have worked with partial fractions a little bit, but I don't see how it applies to this one. I don't mean to second guess you, but are you sure the LHS is not correct? I worked it out again and got the same answer and my TI-89 gives the same answer as well.
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    Idea of how to separate?

    Cool Thanks. So I worked this one out and got the left hand side of the equation, but I can seem to get the integration on the right. \frac{ydy}{(y+1)^2} => \frac{dx}{(1-x^2)} = \int{\frac{ydy}{(y+1)^2}} = \int{\frac{dx}{(1-x^2)}} so, ln|y+1|+\frac{1}{y+1} = \int{\frac{dx}{(1-x^2)}}...
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    Idea of how to separate?

    Ok, so I worked it out some more and got \frac{ydy}{(y+1)^2} = \frac{dx}{(1-x^2)} Is this correct?
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    Idea of how to separate?

    Problem: (y-yx^2)\frac{dy}{dx} = (y+1)^2 So, the first thing I tried was just dividing the whole equation by (y-yx^2) and then factored out the y to get \frac{dy}{dx} = \frac{(y+1)^2}{y(1-x^2)}. Next I expanded the numerator on the right side of the equation and then split them all into...
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    Integral and series

    Oh I see, well, just in case you want to know, the binomial expansion series says you can expand a binomial according to the following formula: 1+kx+\frac{k(k-1)}{2!}x^2+\frac{k(k-1)(k-2)}{3!}x^3 and so on, where k is your exponent. In this integral, \int{\sqrt{x^3 +1}} your function is...
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    Help with a simple Diff Eqn

    Awesome, that was super easy once you look at it that way. Thanks, now I can look at other integrals and apply the same method. Life somehow just became much easier!! ^_^ Thank you!!!
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    Help with a simple Diff Eqn

    I came accross another integral that I am not catching here that seems to be along the same line as this one. Is there a rule to dealing with these kinds of integrals? \int{\frac{x^2}{(1+x)}}dx
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    Help with a simple Diff Eqn

    Here is the problem: Solve, (x+1)\frac{dy}{dx} = x + 6 Here is what I tried: I moved all the x's to one side and left the dy on the left of the equal sign to solve with the separation of variable method. I got, \int{dy} = \int{\frac{(x+6)}{(x+1)}dx} So here I just solve the...
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