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Homework Help: Need some explanation please

  1. Apr 30, 2006 #1
    The spread of a disease in a community is modelled by the following differential equation:

    dy/dx = 0.2y - 0.02x where y is the number of infected individuals in thousands, and x the time in days.

    1) Show the equation that for the family of 'curves' in the x-y plane for which dy/dx is a constant, is of the form y=mx+c.

    How can I do this please?
     
  2. jcsd
  3. Apr 30, 2006 #2

    AKG

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    You have the differential equation:

    y' - 0.2y = p(x)

    where p(x) is some polynomial in x, and y' denotes dy/dx. Find the family of functions that solves this equation, and determine the one that has y' constant.
     
  4. Apr 30, 2006 #3
    p(x) = -0.02x no?
     
  5. Apr 30, 2006 #4
    How is this done :-(
     
  6. Apr 30, 2006 #5

    AKG

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    Yes, that's correct for p(x). To find the general solution y, a function of x, to this problem, you find

    i) the general solution to the related homogeneous equation y' - 0.2y = 0
    ii) find any one particular solution to the original homogenous equation
    iii) add them

    So finding the general solution to y' - 0.2y = 0 means finding a family of solutions to this equation. We can do it:

    y' - 0.2y = 0
    y' = 0.2y
    dy/dx = 0.2y
    dy/y = 0.2dx
    lny = 0.2x + C
    y = exp(0.2x + C)
    y = a*exp(0.2x) [where a = eC]

    So this gives a family of solutions, paramterized by a. In other words, for EACH real number a, y(x) = ae0.2x is a solution to y' - 0.2y = 0. We write yh(x) = ae0.2x, where the subscript "h" stands for "homogenous"

    We now look for yp, a particular solution to the original equation:

    y' - 0.2y = p(x)

    Since p(x) is a degree 1 polynomial, it's a rule of thumb that we ought to guess a degree 1 polynomial for our particular solution. Well a degree 1 polynomial is just one of the form mx+b, so we let this be our guess of yp, and we sub it in to see if it works, and what values of m, b we need:

    If yp = mx+b, then yp' = m, so:

    yp' - 0.2yp = -0.02x
    m - 0.2(mx + b) = -0.02x
    -0.2mx + (m - 0.2b) = -0.02x + 0

    So -0.2m = -0.02 [equating coefficients of x] and m-0.2b = 0 [equating coefficients of 1]

    We get m = 1/10, b = 1/2, giving yp = 0.1x + 0.5

    So the general solution is y = yh + yp = ae0.2x + 0.1x + 0.5. What this means is that a function y of x is a solution to dy/dx = 0.2y - 0.02x if and only if there is some real number a such that y(x) = ae0.2x + 0.1x + 0.5. So if every solution comes in the form ae0.2x + 0.1x + 0.5, which of these have derivative constant?

    y' = 0.2ae0.2x + 0.1

    Well this is constant iff a = 0, correct? And a = 0 iff y = 0.1x + 0.5. So y' is constant iff y = 0.1x + 0.5 which implies that y is in the form y = mx+b.
     
  7. Apr 30, 2006 #6

    nrqed

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    dy/dx is a constant, call it ''k'' let's say. Then replace dy/dx in your initial equation by k and isolate y.
     
  8. Apr 30, 2006 #7

    AKG

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    Yes, do what nrqed said and totally ignore what I said. It's way more work than necessary.
     
  9. Apr 30, 2006 #8
    so

    0.2y = k + 0.02x

    Hence

    y = 0.1x + 5k is this correct?
     
    Last edited: Apr 30, 2006
  10. Apr 30, 2006 #9

    AKG

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    Yes............
     
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