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Thread: ODE help

  1. #1
    tak
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    ODE help

    Code:
    $\displaystyle 
    (3x^2y-y^3)dx-(3xy^2-x^3)dy=0
    $
    Please help solve the above


    $\displaystyle
    d^2y/dx^2 -4dy/dx +3y = e^3e^x
    $

    Please solve both
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  2. #2
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    First one looks exact to me (divide out the $\displaystyle dx$). Then use the procedure involving potential functions for exact ODEs to find the solution. second is a linear non-homogeneous equation, so your general solution will be the general solution of the associated homogeneous equation plus any particular solution of the non-homogeneous equation. Since you have constant coefficients, you should use the method of undetermined coefficients.

    What have you already tried?
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  3. #3
    tak
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    trying out

    I got $\displaystyle xy^3 - x^3y -2/x^2 =a$ where a is a constant
    I use dz=partial z/partial x dx + partial z/ partial y dy method
    dunno correct or not

    by the way what is the definition of homogenous in layman terms thanks
    As for the second Q i got y= C1e^3x +C2e^x
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  4. #4
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    Quote Originally Posted by tak View Post
    I got $\displaystyle xy^3 - x^3y -2/x^2 =a$ where a is a constant
    I use dz=partial z/partial x dx + partial z/ partial y dy method
    dunno correct or not

    by the way what is the definition of homogenous in layman terms thanks
    As for the second Q i got y= C1e^3x +C2e^x
    The homogeneous solution is just what you get if the right side, typically written as $\displaystyle f(x)$ is zero. It is independent of the right side of the fuction, or the inhomogeneous part. In other words, you will get the same homogeneous solution, no matter what $\displaystyle f(x)$ is equal to.

    Edit...It might help to know the standard form for a Second Order ODE is $\displaystyle Ay''+By'+Cy=f(x)$

    Edit 2...This answer is correct so far, but you have only solved for the homogeneous solution. You need to set up, using the method of undetermined coefficients, a solution to the inhomogeneous part. It will be of the form: $\displaystyle Ate^x$ because you need a solution linearly independent of the solutions you have already gotten.
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  5. #5
    tak
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    constant zero

    I found my c3 to be zero; the general solution is just the complementry solution. correct?
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  6. #6
    tak
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    answer correct?

    y=C1e^(3x) +C2xe^(3x) +0.5x^(3x)
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