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Math Help - Very simple differential equation

  1. #1
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    Very simple differential equation

    dy/dt = t^3 * y^3

    This is supposed to be solved using seperation of variables, but after seperating and integrating both sides, you end up with y^-2 on one side and a negative number on the other side (after you multiply both sides by -2 to begin to solve for y). Solving for y would require rasing each side to the -1/2 power... but you can't take the square root of a negative number. That is where I'm stuck.

    Thanks in advance.


    edit: here is what I end up with:

    y^-2 = -1/2 * t^4
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by erorr View Post
    dy/dt = t^3 * y^3

    This is supposed to be solved using seperation of variables, but after seperating and integrating both sides, you end up with y^-2 on one side and a negative number on the other side (after you multiply both sides by -2 to begin to solve for y). Solving for y would require rasing each side to the -1/2 power... but you can't take the square root of a negative number. That is where I'm stuck.

    Thanks in advance.


    edit: here is what I end up with:

    y^-2 = -1/2 * t^4
    This is an SDE

    So we begin by seperating

    \frac{dy}{dt}=t^3\cdot{y^3}\Rightarrow{\frac{dy}{y  ^3}=t^3dt}

    Integrating we get

    \frac{-1}{2y^2}=\frac{t^4}{4}+C

    Solving we get

    \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}


    I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?
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  3. #3
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    Krizalid's Avatar
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    Quote Originally Posted by Mathstud28 View Post

    I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?
    The thing is that you solved wrong the following:

    Quote Originally Posted by Mathstud28 View Post

    Integrating we get

    \frac{-1}{2y^2}=\frac{t^4}{4}+C

    Solving we get

    \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Mathstud28 View Post
    This is an SDE

    So we begin by seperating

    \frac{dy}{dt}=t^3\cdot{y^3}\Rightarrow{\frac{dy}{y  ^3}=t^3dt}

    Integrating we get

    \frac{-1}{2y^2}=\frac{t^4}{4}+C

    Solving we get

    \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}


    I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?
    The C got lost

    \frac{-1}{2y^2}=\frac{t^4}{4}+C

    \frac{-1}{2y^2}=\frac{t^4+C_1}{4}


    y^2=\frac{4}{-2t^4+C_2}

    y=\pm \frac{2}{\sqrt{-2t^4+K}}

    I hope this helps

    Good luck.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    This is an SDE

    So we begin by seperating

    \frac{dy}{dt}=t^3\cdot{y^3}\Rightarrow{\frac{dy}{y  ^3}=t^3dt}

    Integrating we get

    \frac{-1}{2y^2}=\frac{t^4}{4}+\color{red}C

    Solving we get

    \frac{-2}{t^4}=y^2\Rightarrow{y=\pm\frac{i\sqrt{2}}{t^2}}


    I have never seen a diff equation at SDE level with an imaginary part in its answer...maybe you copied it down wrong?
    Don't forget about the C! That will change the answer a bit...it may get rid of the pesky i term.

    y^{-2}=C-\frac{t^4}{2} \implies y=\frac{1}{\sqrt{C-\frac{1}{2}t^4}}
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