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Math Help - Simple Differential Equations Question

  1. #1
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    Simple Differential Equations Question

    I need help finding the solution to this differential equation.

    dz/dt = 9te^(4z)

    I cross multiply so that it becomes: dz/e^(4z) = 9t dt

    From here I need to integrate.

    on the left side, I do substitution and this is what I did:

    u = e^(4z)
    du = 4e^(4z) dz
    du/4e^(4z) = dz

    From here, my equation looked like:

    du/4ue^(4z) = (9t^2)/2 + C
    1/4e^(4z) ln|u| = (9t^2)/2 + C

    I subbed in back the u in the absolute brackets but I am unsure of what to do next. I need to sub the points in f(0) = 0 but if I do using my equation, I am going to get 0, which I think is incorrect. Any ideas of where I went wrong?
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  2. #2
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    Quote Originally Posted by proski117 View Post
    I need help finding the solution to this differential equation.

    dz/dt = 9te^(4z)

    I cross multiply so that it becomes: dz/e^(4z) = 9t dt

    From here I need to integrate.

    on the left side, I do substitution and this is what I did:

    u = e^(4z)
    du = 4e^(4z) dz
    du/4e^(4z) = dz

    From here, my equation looked like:

    du/4ue^(4z) = (9t^2)/2 + C
    1/4e^(4z) ln|u| = (9t^2)/2 + C

    I subbed in back the u in the absolute brackets but I am unsure of what to do next. I need to sub the points in f(0) = 0 but if I do using my equation, I am going to get 0, which I think is incorrect. Any ideas of where I went wrong?
    There is no such thing as "cross multiplying" with derivatives. A derivative is NOT a fraction.

    Anyway

    \frac{dz}{dt} = 9t\,e^{4z}

    e^{-4z}\,\frac{dz}{dt} = 9t

    \int{e^{-4z}\,\frac{dz}{dt}\,dt} = \int{9t\,dt}

    \int{e^{-4z}\,dz} = \frac{9t^2}{2} + C_1

    -\frac{1}{4}e^{-4z} + C_2 = \frac{9t^2}{2} + C_1

    -\frac{1}{4}e^{-4z} = \frac{9t^2}{2} + C_1 - C_2

    e^{-4z} = -18t^2 + C where C = -4(C_1 - C_2)

    -4z = \ln{|C - 18t^2|}

    z = -\frac{1}{4}\ln{|C - 18t^2|}.


    Do you have a boundary condition so that you can find the unknown?
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  3. #3
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    it says the point of origin
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  4. #4
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    Then substitute z = 0 and t = 0 and solve for C.
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  5. #5
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    by subbing in z,t = 0, you are left with:

    -1/4 ln|c| = 0

    what do you do from here? is your c = -1/4?
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  6. #6
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    Quote Originally Posted by proski117 View Post
    by subbing in z,t = 0, you are left with:

    -1/4 ln|c| = 0

    what do you do from here? is your c = -1/4?
    No.

    Go to the step that doesn't have a modulus sign in it...

    e^{-4z} = -18t^2 + C

    Substitute the values

    e^{-4(0)} = -18(0)^2 + C

    1 = C.


    So that means

    z = -\frac{1}{4}\ln{(1 - 18t^2)}.


    I just realised too that you don't need the modulus signs at all...
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  7. #7
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    it says i am trying to look for the z value, and i entered -1/4 and it was incorrect

    Edit: I think i am supposed to enter the equation, not the value of C. Thank you very much for your help.
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