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Calculus

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Find the complete solution to the 2nd ODE:

2y''+3y'+y=t^(2)+3sin(t)

  • Calculus-ODE -

    Start with solving for the complementary solution to the homogeneous equation
    2y"+3y'+y=0,
    the characteristic equation is
    2r²+3r+1=0
    which is an algebraic equation which can be factorized into
    (2r+1)(r+1)=0
    or
    r={-(1/2), -1}
    The complementary solution is therefore:
    yc=Ae-(x/2) + Be-x

    The next step is to find the particular solution for the right-hand side:
    y²+3sin(t)

    For this, we will need to assume that the particular solution yp is as follows:
    yp=Ct²+Dt+Esin(t)+Fcos(t)+G

    Now to find the constants C,D,E,F,G, we need to calculate
    2yp"+3yp'+yp=t²+sin(t)
    and equate coefficients.
    We end up with:
    G+(-3sin(t)-cos(t))F+(3cos(t)-sin(t))E+(t+3)D+(t^2+6*t+4)C = t²+sin(t)
    which can solved for C,D,E,F,G by comparing coefficients and successive substitution:

    Sum coefficients of sin(t)=1,
    sum of coefficients of cos(t)=0:
    -3F-E=1
    -F+3E=0
    which when solved, gives
    E=-1/10, F=-3/10.

    Now tackle C, D and G:
    Compare coefficients of t² yields
    C=1
    Substitute C and compare coefficients of t gives D=-6
    and consequently coefficient G=14.

    The general solution is therefore:
    y(t)=yc+yp
    =Ae-(t/2) + Be-t +
    t²-6t -sin(t)/10 -3cos(t)/10+14

    Finally, we check that
    2y"(t)+3y'(t)+y(t) equals t²+sin(t)
    which confirms that the solution is correct.

  • Calculus -

    If you need help with finding the standard forms of the particular solution, check your course notes, your textbook, or the following link:

    http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

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