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Math Help - 2nd order differential

  1. #1
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    2nd order differential

    can anyone give msom light on this question, thanks

    find the complete solution of the differential equation
    d2y/dx2 4 dy/dx + 3y = 65 sin 2 x

    find the value of the constant a such that y= axe^-1 is a solution of the differential equation
    d2y/dx2 + 3 dy/dx + 2y = 2 e-x
    find the solution for which y=1 and dy/dx =3 when x=0
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  2. #2
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    Quote Originally Posted by divine_mail View Post
    can anyone give msom light on this question, thanks

    find the complete solution of the differential equation
    d2y/dx2 – 4 dy/dx + 3y = 65 sin 2 x

    Mr F says: It's expected that you can find the homogeneous solution, that is, the solution to {\color{red}\frac{d^2 y}{dx^2} - 4 \frac{dy}{dx} + 3y = 0}.

    So I'll assume that your trouble is in getting a particular solution.

    It's unclear from your notation whether you mean {\color{red}65 \sin (2x)} or {\color{red}65 \sin^2 x}.

    If {\color{red}65 \sin (2x)} then try {\color{red} y = a \sin (2x) + b \cos (2x)} as a particular solution.

    If {\color{red}65 \sin^2 x} then note that {\color{red}65 \sin^2 x = \frac{65}{2} (1 - \cos (2x)) = \frac{65}{2} - \frac{65}{2} \cos (2x)} and so you should try {\color{red} y = a \sin (2x) + b \cos (2x) + c} as a particular solution.

    In each case you substitute the particular solution into the DE and solve for the values of a, b, c .......

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    find the value of the constant a such that y= axe^-1 is a solution of the differential equation Mr F says: You mean y= axe^-x.

    d2y/dx2 + 3 dy/dx + 2y = 2 e-x


    Mr F says: You're being asked this question because it's the form of a particular solution to the DE.

    The general solution to the DE is the homogeneous solution plus the particular solution.

    It's expected that you can find the homogeneous solution. It's {\color{red} y = A e^{-x} + B e^{-2x}}.

    Then the general solution is {\color{red} y = A e^{-x} + B e^{-2x} + a x e^{-x}}. The value of a is found below.


    find the solution for which y=1 and dy/dx =3 when x=0

    Mr F says: Substitute these boundary conditions into the general solution to get the value of A and B.
    y = a x e^{-x} therefore:

    \frac{dy}{dx} = a e^{-x} - ax e^{-x} = a e^{-x} (1 - x).

    \frac{d^2y}{dx^2} = - a e^{-x} (1 - x) - a e^{-x} = -a e^{-x} (2 - x).


    Substitute these expressions into the DE and simplify:


    -a e^{-x} (2 - x) + 3 a e^{-x} (1 - x) + 2 a x e^{-x} = 2 e^{-x}

    \Rightarrow -a (2 - x) + 3a (1 - x) + 2ax = 2

    \Rightarrow a = 2.
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