differential equation??

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  • 4 weeks ago

    I assume you want the solution?

    3y" + 2y' + y = 0

    let y = e^(rx) then the ODE transforms into

    (3r^2 + 2r + 1)e^(rx) =0 = (3r^2 + 2r + 1)

    Solve the quadratic for r:

    r = -1/3 +/- (1/6)*sqrt((2)^2 -4*3) = (1/3)*{ -1 +/- (1/2) sqrt(-2)} =(1/3){ -1 +/-  i/sqrt(2)}

    So

    y = e^(-x/3)*{A*e^(i x/(3*sqrt(2))) + B*e^(-i x/(3*sqrt(2))) }

    A and B are determined from boundary or initial conditions

  • Fuhr
    Lv 6
    4 weeks ago

    Why yes.  Yes it is.

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