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October 4 Math 2306 sec. 51 Fall 2019 Section 9: Method of Undetermined Coefficients The context here is linear, constant coefficient, nonhomogeneous equations a n y (n) + a n-1 y (n-1) + ··· + a 0 y = g (x ) where g comes from the restricted classes of functions I polynomials, I exponentials, I sines and/or cosines, I and products and sums of the above kinds of functions Recall y = y c + y p , so we’ll have to find both the complementary and the particular solutions! October 2, 2019 1 / 33

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  • October 4 Math 2306 sec. 51 Fall 2019

    Section 9: Method of Undetermined Coefficients

    The context here is linear, constant coefficient, nonhomogeneousequations

    any (n) + an−1y (n−1) + · · ·+ a0y = g(x)

    where g comes from the restricted classes of functionsI polynomials,I exponentials,I sines and/or cosines,I and products and sums of the above kinds of functions

    Recall y = yc + yp, so we’ll have to find both the complementary andthe particular solutions!

    October 2, 2019 1 / 33

  • Motivating ExampleFind a particular solution of the ODE

    y ′′ − 4y ′ + 4y = 8x + 1

    Question: What kind of function yp(x) when plugged into the rightside might equal a first degree polynomial? Perhaps another firstdegree polynomial.

    Make an educated guess that

    yp(x) = Ax + B for some numbers A and B.

    October 2, 2019 2 / 33

  • October 2, 2019 3 / 33

  • October 2, 2019 4 / 33

  • The Method: Assume yp has the same form as g(x)

    y ′′ − 4y ′ + 4y = 6e−3x

    Question: What kind of function yp(x) when plugged into the rightside might equal a multiple of e−3x?

    Make an educated guess that

    yp(x) = Ae−3x for some number A.

    October 2, 2019 5 / 33

  • October 2, 2019 6 / 33

  • Make the form general

    y ′′ − 4y ′ + 4y = 16x2

    There are two ways to view the right side g(x) = 16x2, as a monomialAx2 or as a quadratic expression. Let’s try assuming that

    yp(x) = Ax2

    October 2, 2019 7 / 33

  • October 2, 2019 8 / 33

  • October 2, 2019 9 / 33

  • October 2, 2019 10 / 33

  • General Form: sines and cosines

    y ′′ − y ′ = 20 sin(2x)

    If we assume that yp = A sin(2x), taking two derivatives would lead tothe equation

    −4A sin(2x)− 2A cos(2x) = 20 sin(2x).

    This would require (matching coefficients of sines and cosines)

    −4A = 20 and − 2A = 0.

    This is impossible as it would require −5 = 0!

    October 2, 2019 11 / 33

  • General Form: sines and cosines

    We must think of our equation y ′′ − y ′ = 20 sin(2x) as

    y ′′ − y ′ = 20 sin(2x) + 0 cos(2x).

    The correct format for yp is

    yp = A sin(2x) + B cos(2x).

    October 2, 2019 12 / 33

  • Examples of Forms of yp based on g (Trial Guesses)

    (a) g(x) = 1 (constant function)

    (b) g(x) = x − 7 (1st degree polynomial)

    October 2, 2019 13 / 33

  • Examples of Forms of yp based on g (Trial Guesses)

    (c) g(x) = 5x2 (2nd degree polynomial)

    (d) g(x) = 3x3 − 5 (3rd degree polynomial)

    October 2, 2019 14 / 33

  • Examples of Forms of yp based on g (Trial Guesses)

    (e) g(x) = xe3x (1st degree polynomial times e3x )

    (f) g(x) = cos(7x) (sum of cos(7x) and sin(7x))

    October 2, 2019 15 / 33

  • Examples of Forms of yp based on g (Trial Guesses)

    (g) g(x) = sin(2x)− cos(4x)(sum of cos(2x), sin(2x), cos(4x) and sin(4x))

    (h) g(x) = x2 sin(3x)(2nd degree poly. times sin(3x) and 2nd degree poly. times cos(3x))

    October 2, 2019 16 / 33

  • Examples of Forms of yp based on g (Trial Guesses)

    (i) g(x) = ex cos(2x) (sum of ex cos(2x) and ex sin(2x))

    (j) g(x) = xe−x sin(πx)(sum of 1st deg. poly. times ex cos(πx) and 1st deg. poly. timesex sin(πx))

    October 2, 2019 17 / 33