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Math Help - Differential equation with initial conditions

  1. #1
    Newbie Exotique's Avatar
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    Differential equation with initial conditions

    Find the unique function y(x) satisfying the differential equation with initial condition,

    \frac{dy}{dx}=x^2 y, y(1)=1
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Exotique View Post
    Find the unique function y(x) satisfying the differential equation with initial condition,

    \frac{dy}{dx}=x^2 y, y(1)=1
    \int \frac{dy}{y} = \int x^2 dx

    \ln{|y|} = \frac{x^3}{3} + C

    Can you take it fron here?
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  3. #3
    Newbie Exotique's Avatar
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    I'm guessing I have to solve for y in terms of x, let y = 1 and x = 1, then solve for C?

    This is what I did:

    y = e^\frac{x^3}{3} + C

    Since y(1)=1,

    1=e^\frac{1}{3} + C

    C=1 - e^\frac{1}{3}

    For some reason this doesn't feel right.

    Then the original equation should be:

    y = e^\frac{x^3}{3} + 1 - e^\frac{1}{3}
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Exotique View Post
    I'm guessing I have to solve for y in terms of x, let y = 1 and x = 1, then solve for C?

    This is what I did:

    y = e^\frac{x^3}{3} + C

    Since y(1)=1,

    1=e^\frac{1}{3} + C

    C=1 - e^\frac{1}{3}

    For some reason this doesn't feel right.


    Then the original equation should be:

    y = e^\frac{x^3}{3} + 1 - e^\frac{1}{3}
    y= e^{\frac{x^3}{3}+ C}

    y = e^\frac{x^3}{3}e^C

    y=Ce^{\frac{x^3}{3}}

    1=Ce^{\frac{1}{3}}

    \frac{1}{e^{\frac{1}{3}}} = C
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  5. #5
    Newbie Exotique's Avatar
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    How did you get from
    y=e^\frac{x^3}{3} e^C

    to
    y=Ce^\frac{x^3}{3}
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  6. #6
    Master Of Puppets
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    e to the power of an arbitrary constant remains an arbitrary constant. Therefore we can say e^C = C
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  7. #7
    Newbie Exotique's Avatar
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    Ahh, I understand now, thank you to all.
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  8. #8
    Flow Master
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    Quote Originally Posted by Exotique View Post
    Ahh, I understand now, thank you to all.
    This question counts towards the student's final grade. It is not MHF policy to knowingly assist with such questions. Thread closed.
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