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Math Help - solving differential equation using substution

  1. #1
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    solving differential equation using substution

    DY/DX = (7X - 3Y)^3

    I used substitution Z = (7X -3Y) Then seperate the variables but i cannot do the integration of 1/(Z^3 - 7) Unless i am supposed to use another substitution??

    Thanks!
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  2. #2
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    Quote Originally Posted by Niall101 View Post
    DY/DX = (7X - 3Y)^3

    I used substitution Z = (7X -3Y) Then seperate the variables but i cannot do the integration of 1/(Z^3 - 7) Unless i am supposed to use another substitution??

    Thanks!
    If you are using the substitution z = 7x - 3y, note that

    \frac{dz}{dx} = \frac{d}{dx}(7x - 3y)

    \frac{dz}{dx} = 7 - 3\frac{dy}{dx}.


    So \frac{dy}{dx} = \frac{1}{3}\left(7 - \frac{dz}{dx}\right).


    So from your original DE

    \frac{dy}{dx} = (7x - 3y)^3

    \frac{1}{3}\left(7 - \frac{dz}{dx}\right) = z^3

    7 - \frac{dz}{dx} = 3z^3

    \frac{dz}{dx} = 7 - 3z^3

    \frac{dx}{dz} = \frac{1}{7 - 3z^3}.


    You can now solve this using partial fractions.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The roots of the polynomial z^{3}- 7 are...

    z_{0} = \sqrt[3]{7}

    z_{1} = \sqrt[3]{7}\cdot e^{i \frac{2}{3} \pi}

    z_{2} = \sqrt[3]{7}\cdot e^{i \frac{4}{3} \pi} (1)

    ... so that is...

    \frac{1}{z^{3}-7} = \frac{a_{0}}{z-z_{0}} + \frac{a_{1}}{z-z_{1}} + \frac{a_{2}}{z-z_{2}} (2)

    ... where...

    a_{k} = \lim_{z \rightarrow z_{k}} \frac{z-z_{k}}{z^{3} - 7}, k=0,1,2 (3)

    Kind regards

    \chi \sigma
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    Quote Originally Posted by Prove It View Post
    If you are using the substitution z = 7x - 3y, note that

    \frac{dz}{dx} = \frac{d}{dx}(7x - 3y)

    \frac{dz}{dx} = 7 - 3\frac{dy}{dx}.


    So \frac{dy}{dx} = \frac{1}{3}\left(7 - \frac{dz}{dx}\right).


    So from your original DE

    \frac{dy}{dx} = (7x - 3y)^3

    \frac{1}{3}\left(7 - \frac{dz}{dx}\right) = z^3

    7 - \frac{dz}{dx} = 3z^3

    \frac{dz}{dx} = 7 - 3z^3

    \frac{dx}{dz} = \frac{1}{7 - 3z^3}.


    You can now solve this using partial fractions.
    Hey thank you! I had this on one of my attempts with disregarded it as i ended up with a horrible looking solution. Thatnks so much!
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