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A Nonlinear Second Order Differential Equation Problem

  1. Nov 7, 2009 #1
    A Nonlinear Second Order Differential Equation Problem: very frustrating please help!

    Hello, I am a first year engineering undergraduate student, and this is my question.

    1. The problem statement, all variables and given/known data
    A dust particle of negligible mass starts to fall, t=0, under the influence of gravitational force through mist of saturated water vapor. The vapor condenses onto the dust particle at a constant rate of [tex]\lambda[/tex] kilogram per meter of traveled distance. The dust particle thereby develops into a rain droplet with increasing mass.

    a) Calculate the acceleration of the droplet as a function of its velocity and covered distance.

    b) Determine the equation of motion of the droplet by integrating the expression of the acceleration, neglecting friction, collisions, etc.


    2. Relevant equations
    I have already solved a). The equation (which I am told is correct) is:
    [tex]\frac{dv}{dt} + \frac{v^{2}}{x} = [/tex] g

    where g is the gravitational field strength.

    3. The attempt at a solution
    I was given the 'hint' of substituting [tex]f(x) = At^{n}[/tex] to solve this equation for the position x, and determine A and n.

    I have no idea how to determine A and n, frankly. In fact, I have never done Second Order ODEs before, let alone nonlinear Second Order ODEs. (Before you ask why I would be set a question on Second Order ODEs not having done it before - my professor - let's just say I've had better elementary schools teachers.)

    I arrive at this equation:

    (2n[tex]^{2}-n)At^{n-2} = [/tex] g

    What do I do from here now? Four hours of rearranging to no avail.

    P.S. After sleuthing around, I discovered that the usual method for solving such equations is to substitute it to a first order one. This doesn't look like it...
     
    Last edited: Nov 7, 2009
  2. jcsd
  3. Nov 8, 2009 #2
    bump =(
     
  4. Nov 8, 2009 #3

    ehild

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    Homework Helper
    Gold Member

    Hi Overmage,

    Your equation

    [tex]
    \frac{dv}{dt} + \frac{v^{2}}{x} = g
    [/tex]

    is correct, it was a good job to derive it!


    Now it is a first order differential equation (contains first derivative only), but non-linear. Luckily, it does not contain the time itself. There is a trick for such equations, often applied in Mechanics: We will use x as independent variable instead of t.

    [tex]
    \frac{dv}{dt} =\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}
    [/tex]

    Substituting for dv/dt, your equation becomes:

    [tex]
    \frac{dv}{dx}v+\frac{v^{2}}{x} = g
    [/tex]

    Now the second trick comes:

    [tex]
    \frac{dv}{dx}v=0.5\frac{d(v^2)}{dx}
    [/tex]

    and we choose v^2 as dependent variable, say y=v^2.

    Now we arrived at a totally different-looking equation:

    [tex]
    0.5 \frac{dy}{dx}+\frac{y}{x} = g
    [/tex]

    This is a linear equation! You can solve it by assuming y in a product form: y=F*G.

    Have you learnt that method from Maths? If you do not know it yet, I'll explain.

    Anyway: the solution will be:

    [tex]
    v^2=\frac{2gx}{3}
    [/tex]

    from where you get the acceleration, and it will be a surprise!

    ehild
     
  5. Nov 8, 2009 #4
    No, I'm not familiar with that. =( could you give me a general idea of how that works out?
     
  6. Nov 9, 2009 #5

    ehild

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    Homework Helper
    Gold Member

    You have a differential equation of form

    a(x)y'+b(x)y=c(x) (the prime means differentiation), with initial condition y=0 at x=0, the prime means differentiation.

    Assume Y=F(X)G(x).

    a(x)(F'G+FG')+b(x)FG =c(x). *

    You have the freedom to choose G so that

    a(x)FG' + b(x) FG =0

    Divide the eq. by F.

    a(x)G'+b(X)G=0.

    You can solve this equation by separating the variables.

    [tex] \int{\frac{dG}{G}}=\int{\frac{-b(x)}{a(x)}dx} [/tex]

    Replace back G(x) into eq. *.

    a(x)F'G(x) = c(x)

    [tex]F(x)=\int{\frac{c(x)}{a(x)G(x)}}+const[/tex]

    y=F(x)G(x).
    -----------------------------------------------------------------------

    Your equation is :

    0.5 y' +y/x =g ---> y'+2y/x=2g

    a=1, b=2/x, c=g


    [tex] \int{\frac{dG}{G}}=-\int{\frac{2}{x}dx} \rightarrow G(x)=x^{-2}[/tex]

    [tex] F'G(x)=c(x)\rightarrow F'=2gx^2 \rightarrow

    F(x)=\frac{2gx^3}{3}+C[/tex]

    [tex]y(x)= F(x)G(x)=\frac{2gx}{3}+\frac{C}{x^2} [/tex]

    C=0 according to the initial condition.

    [tex]y(x)=\frac{2gx}{3}[/tex]

    ehild
     
  7. Nov 9, 2009 #6
    Hey,

    Thanks alot for the help! I really appreciate it. However, can i clarify just one last thing:

    shouldn't c be = 2g?

    Thanks for the help you've given! Wow, this problem is kinda hard. Think you've pretty much covered it all though.
     
    Last edited: Nov 9, 2009
  8. Nov 9, 2009 #7

    ehild

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    Homework Helper
    Gold Member

    Yes, of course... :)


    ehild
     
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