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Homework Help: Rawr! Stuck on ODE (linear 1st order!)

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve the initial value problem:

    y'+[itex]\frac{4y}{x+8}[/itex]=(x+8)[itex]^{8}[/itex] , y(0)=8.

    The differential equation is linear.

    2. Relevant equations


    3. The attempt at a solution

    I can see that the equation is in the form y' +P(x)*y = Q(x) so I'm like "easy, lemme get an integration factor going and solve this mofo."

    I= e[itex]^{\int P(x)dx}[/itex] = e[itex]^{\int \frac{4}{x+8}dx}[/itex] = e[itex]^{4 ln |x+8|}[/itex] = (x+8)[itex]^{4}[/itex] (I took the positive expression since any number raised to the 4th power is positive...)

    Then multiply the equation by I:

    (x+8)[itex]^{4}[/itex]*y' + (x+8)[itex]^{4}[/itex]*4(x+8)[itex]^{-1}[/itex]*y = (x+8)[itex]^{4}[/itex]*(x+8)[itex]^{8}[/itex]


    y'(x+8)[itex]^{4}[/itex]+4y(x+8)[itex]^{3}[/itex] = (x+8)[itex]^{12}[/itex]

    Now the LHS is the derivative w/ respect to x for the product y*(x+8)[itex]^{4}[/itex] and by integrating both sides I get:

    y(x+8)[itex]^{4}[/itex] = [itex]\frac{(x+8)^{13}}{13}[/itex] + C

    So now I solve for C given the initial conditions and I get a big *** negative number

    C = [itex]\frac{104-8^{9}}{13}[/itex] = -[itex]\frac{134217624}{13}[/itex]

    My final solution looks like this (gotten it 3 times now, so I must be doing the same mistake over and over...)

    y = [itex]\frac{(x+8)^{9} - 134217624}{13}[/itex]

    I feel an almost instinctual compulsion to reject any answer that is ginormous or just crazy looking when it comes to homework problems.. but I submitted anyway and sure enough I'm incorrect. Anyone got a bone to throw for me?
  2. jcsd
  3. Sep 3, 2011 #2
    y(x+8)[itex]^{4}[/itex] = [itex]\frac{(x+8)^{13}}{13}[/itex] + C

    Solve for y here. Don't plug in the initial condition right away.
  4. Sep 3, 2011 #3
    You integrated too soon, write out the dy/dx.
  5. Sep 3, 2011 #4
    I didn't think it'd matter... I get the same thing.

    y = [itex]\frac{(x+8)^{9}}{13}[/itex] + C[itex]_{1}[/itex] | where C[itex]_{1}[/itex] is just another arbitrary constant that will satisfy the equation given the initial conditions. I get the same constant regardless...

    8 = [itex]\frac{8^{9}}{13}[/itex] + C[itex]_{1}[/itex]
    C[itex]_{1}[/itex] = 8 - [itex]\frac{8^{9}}{13}[/itex] = -[itex]\frac{134217624}{13}[/itex]
  6. Sep 3, 2011 #5
    Ok lemme look at that, thanks.
  7. Sep 4, 2011 #6
    No luck. Oh well, thanks for your time I guess.
  8. Sep 4, 2011 #7
    That's such a ridiculous question, who came up with it?

    I keep getting something different than you, a really big number. I think you're supposed to divide the C by the (x+8)^4. You can't change that to another constant because (x+8)^4 isn't a constant. I keep getting -4.22889088e10. Kinda of a ridiculous C and problem.
  9. Sep 5, 2011 #8


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    Homework Helper

    You're fine up to here. But for y(0) = 8 , shouldn't this be

    [tex]8 \cdot 8^{4} = \frac{8^{13}}{13} + C [/tex]

    [tex]\Rightarrow 8^{5} = \frac{8^{13}}{13} + C ,[/tex]

    and so

    [tex]C = 8^{5} - \frac{8^{13}}{13} ?[/tex]

    [Note: you are not obligated to simplify this -- in some situations, it isn't even desirable to do so.]

    Your solution function is then

    [tex] y = \frac{(x+8)^{9}}{13} + [ 8^{5} - \frac{8^{13}}{13} ] \cdot \frac{1}{(x+8)^{4}} . [/tex]

    [In your entry above, don't forget that the "arbitrary constant" also getting divided by (x+8)4.]

    It is pretty easy to see, in this form, that your initial condition is satisfied.

    As far as how you'd come up with such a strange differential equation and solution function, this can arise from the sort of "mixing problems" involving finding the mass function of a substance where solutions of different concentrations are being added and drained from a tank (in which they are "instantaneously mixed"), with the inflow rate being greater than the drainage rate.
    Last edited: Sep 5, 2011
  10. Sep 6, 2011 #9
    Thank you so much guys, I was banging my head on my desk with that. The solution

    y = [itex]\frac{(x+8)^{9}}{13}[/itex] + (8[itex]^{5}[/itex] - [itex]\frac{8^{13}}{13}[/itex])(x+8)[itex]^{-4}[/itex]

    checked out.
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