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Mathematics LibreTexts

5.5E: The Method of Undetermined Coefficients II (Exercises)

  • Page ID
    18288
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    In Exercises [exer:5.5.1} –[exer:5.5.17} find a particular solution.

    [exer:5.5.1] \(y''+3y'+2y=7\cos x-\sin x\)

    [exer:5.5.2] \(y''+3y'+y=(2-6x)\cos x-9\sin x\)

    [exer:5.5.3] \(y''+2y'+y=e^x(6\cos x+17\sin x)\)

    [exer:5.5.4] \(y''+3y'-2y=-e^{2x}(5\cos2x+9\sin2x)\)

    [exer:5.5.5] \(y''-y'+y=e^x(2+x)\sin x\)

    [exer:5.5.6] \(y''+3y'-2y=e^{-2x}\left[(4+20x)\cos 3x+(26-32x)\sin 3x\right]\)

    [exer:5.5.7] \(y''+4y=-12\cos2x-4\sin2x\)

    [exer:5.5.8] \(y''+y=(-4+8x)\cos x+(8-4x)\sin x\)

    [exer:5.5.9] \(4y''+y=-4\cos x/2-8x\sin x/2\)

    [exer:5.5.10] \(y''+2y'+2y=e^{-x}(8\cos x-6\sin x)\)

    [exer:5.5.11] \(y''-2y'+5y=e^x\left[(6+8x)\cos 2x+(6-8x)\sin2x\right]\)

    [exer:5.5.12] \(y''+2y'+y=8x^2\cos x-4x\sin x\)

    [exer:5.5.13] \(y''+3y'+2y=(12+20x+10x^2)\cos x+8x\sin x\)

    [exer:5.5.14] \(y''+3y'+2y=(1-x-4x^2)\cos2x-(1+7x+2x^2)\sin2x\)

    [exer:5.5.15] \(y''-5y'+6y=-e^x\left[(4+6x-x^2)\cos x-(2-4x+3x^2)\sin x\right]\)

    [exer:5.5.16] \(y''-2y'+y=-e^x\left[(3+4x-x^2)\cos x+(3-4x-x^2)\sin x\right]\)

    [exer:5.5.17] \(y''-2y'+2y=e^x\left[(2-2x-6x^2)\cos x+(2-10x+6x^2)\sin x\right]\)

    [exer:5.5.18] \(y''+2y'+y=e^{-x}\left[(5-2x)\cos x-(3+3x)\sin x\right]\)

    [exer:5.5.19] \(y''+9y=-6\cos 3x-12\sin 3x\)

    [exer:5.5.20] \(y''+3y'+2y=(1-x-4x^2)\cos2x-(1+7x+2x^2)\sin2x\)

    [exer:5.5.21] \(y''+4y'+3y=e^{-x}\left[(2+x+x^2)\cos x+(5+4x+2x^2)\sin x\right]\)

    [exer:5.5.22] \(y''-7y'+6y=-e^x(17\cos x-7\sin x), \quad y(0)=4,\; y'(0)=2\)

    [exer:5.5.23] \(y''-2y'+2y=-e^x(6\cos x+4\sin x), \quad y(0)=1,\; y'(0)=4\)

    [exer:5.5.24] \(y''+6y'+10y=-40e^x\sin x, \quad y(0)=2,\quad y'(0)=-3\)

    [exer:5.5.25] \(y''-6y'+10y=-e^{3x}(6\cos x+4\sin x), \quad y(0)=2,\quad y'(0)=7\)

    [exer:5.5.26] \(y''-3y'+2y=e^{3x}\left[21\cos x-(11+10x)\sin x\right], \; y(0)=0, \quad y'(0)=6\)

    [exer:5.5.27] \(y''-2y'-3y=4e^{3x}+e^x(\cos x-2\sin x)\)

    [exer:5.5.28] \(y''+y=4\cos x-2\sin x+xe^x+e^{-x}\)

    [exer:5.5.29] \(y''-3y'+2y=xe^x+2e^{2x}+\sin x\)

    [exer:5.5.30] \(y''-2y'+2y=4xe^x\cos x+xe^{-x}+1+x^2\)

    [exer:5.5.31] \(y''-4y'+4y=e^{2x}(1+x)+e^{2x}(\cos x-\sin x)+3e^{3x}+1+x\)

    [exer:5.5.32] \(y''-4y'+4y=6e^{2x}+25\sin x, \quad y(0)=5,\; y'(0)=3\)

    [exer:5.5.33] \(y''+4y=-e^{-2x}\left[(4-7x)\cos x+(2-4x)\sin x\right], \; y(0)=3, \quad y'(0)=1\)

    [exer:5.5.34] \(y''+4y'+4y=2\cos2x+3\sin2x+e^{-x}, \quad y(0)=-1,\; y'(0)=2\)

    [exer:5.5.35] \(y''+4y=e^x(11+15x)+8\cos2x-12\sin2x, \quad y(0)=3,\; y'(0)=5\)

    [exer:5.5.36]

    Verify that if

    \[y_p=A(x)\cos\omega x+B(x)\sin\omega x\]

    where \(A\) and \(B\) are twice differentiable, then

    \[\begin{aligned} y_p'&=&(A'+\omega B)\cos\omega x+(B'-\omega A) \sin\omega x\mbox{ and}\\ y_p''&=&(A''+2\omega B'-\omega^2A)\cos\omega x +(B''-2\omega A'-\omega^2B)\sin\omega x.\end{aligned}\]

    Use the results of

    a

    to verify that

    \[\begin{aligned} ay_p''+by_p'+cy_p&=&\left[(c-a\omega^2)A+b\omega B+2a\omega B'+bA'+aA''\right] \cos\omega x+\

    \[5pt] && \left[-b\omega A+(c-a\omega^2)B-2a\omega A'+bB'+aB''\right]\sin\omega x.\end{aligned}\]

    Use the results of

    a

    to verify that

    \[y_p''+\omega^2 y_p=(A''+2\omega B')\cos\omega x+ (B''-2\omega A')\sin\omega x.\]

    Prove Theorem [thmtype:5.5.2}.

    [exer:5.5.37] Let \(a\), \(b\), \(c\), and \(\omega\) be constants, with \(a\ne0\) and \(\omega>0\), and let

    \[P(x)=p_0+p_1x+\cdots+p_kx^k \mbox{\quad and \quad} Q(x)=q_0+q_1x+\cdots+q_kx^k,\]

    where at least one of the coefficients \(p_k\), \(q_k\) is nonzero, so \(k\) is the larger of the degrees of \(P\) and \(Q\).

    Show that if \(\cos\omega x\) and \(\sin\omega x\) are not solutions of the complementary equation

    \[ay''+by'+cy=0,\]

    then there are polynomials

    \[A(x)=A_0+A_1x+\cdots+A_kx^k \mbox{\quad and \quad} B(x)=B_0+B_1x+\cdots+B_kx^k \eqno{\rm (A)}\]

    such that

    \[\begin{array}{lcl} \quad (c-a\omega^2)A+b\omega B+2a\omega B'+bA'+aA''&=&P\phantom{.}\

    \[5pt] -b\omega A+(c-a\omega^2)B-2a\omega A'+bB'+aB''&=&Q, \end{array}\]

    where \((A_k,B_k)\), \((A_{k-1},B_{k-1})\), …,\((A_0,B_0)\) can be computed successively by solving the systems

    \[\begin{array}{lcl} \phantom{-}(c-a\omega^2)A_k+b\omega B_k&=&p_k\phantom{.}\

    \[5pt] -b\omega A_k+(c-a\omega^2)B_k&=&q_k, \end{array}\]

    and, if \(1\le r\le k\),

    \[\begin{array}{lcl} \phantom{-}(c-a\omega^2)A_{k-r}+b\omega B_{k-r}&=&p_{k-r}+\cdots\phantom{.}\

    \[5pt] -b\omega A_{k-r}+(c-a\omega^2)B_{k-r}&=&q_{k-r}+\cdots, \end{array}\]

    where the terms indicated by “\(\cdots\)” depend upon the previously computed coefficients with subscripts greater than \(k-r\). Conclude from this and Exercise [exer:5.5.36}

    b

    that

    \[y_p=A(x)\cos\omega x+B(x)\sin\omega x \eqno{\rm (B)}\]

    is a particular solution of

    \[ay''+by'+cy=P(x)\cos\omega x+Q(x)\sin\omega x.\]

    Conclude from Exercise [exer:5.5.36}

    c

    that the equation

    \[a(y''+\omega^2 y)=P(x)\cos\omega x+Q(x)\sin\omega x \eqno{\rm (C)}\]

    does not have a solution of the form (B) with \(A\) and \(B\) as in (A). Then show that there are polynomials

    \[A(x)=A_0x+A_1x^2+\cdots+A_kx^{k+1}\mbox{\quad and \quad} B(x)=B_0x+B_1x^2+\cdots+B_kx^{k+1}\]

    such that

    \[\begin{array}{rcl} a(A''+2\omega B')&=&P\

    \[5pt] a(B''-2\omega A')&=&Q, \end{array}\]

    where the pairs \((A_k,B_k)\), \((A_{k-1},B_{k-1})\), …, \((A_0,B_0)\) can be computed successively as follows:

    \[\begin{aligned} A_k&=&-{q_k\over2a\omega(k+1)}\

    \[5pt] B_k&=&\phantom{-}{p_k\over2a\omega(k+1)},\end{aligned}\]

    and, if \(k\ge 1\),

    \[\begin{aligned} A_{k-j}&=&-{1\over2\omega} \left[{q_{k-j}\over a(k-j+1)}-(k-j+2)B_{k-j+1}\right]\

    \[5pt] B_{k-j}&=&\phantom{-}{1\over2\omega} \left[{p_{k-j}\over a(k-j+1)}-(k-j+2)A_{k-j+1}\right]\end{aligned}\]

    for \(1\le j\le k\). Conclude that (B) with this choice of the polynomials \(A\) and \(B\) is a particular solution of (C).

    [exer:5.5.38] Show that Theorem [thmtype:5.5.1} implies the next theorem: Suppose \(\omega\) is a positive number and \(P\) and \(Q\) are polynomials. Let \(k\) be the larger of the degrees of \(P\) and \(Q\). Then the equation

    \[ay''+by'+cy=e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)\]

    has a particular solution

    \[y_p=e^{\lambda x}\left(A(x)\cos\omega x+B(x)\sin\omega x\right), \eqno{\rm (A)}\]

    where

    \[A(x)=A_0+A_1x+\cdots+A_kx^k \mbox{\quad and \quad} B(x)=B_0+B_1x+\cdots+B_kx^k,\]

    provided that \(e^{\lambda x}\cos\omega x\) and \(e^{\lambda x}\sin\omega x\) are not solutions of the complementary equation. The equation

    \[a\left[y''-2\lambda y'+(\lambda^2+\omega^2)y\right]= e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)\]

    \((\)for which \(e^{\lambda x}\cos\omega x\) and \(e^{\lambda x}\sin\omega x\) are solutions of the complementary equation\()\) has a particular solution of the form (A), where

    \[A(x)=A_0x+A_1x^2+\cdots+A_kx^{k+1} \mbox{\quad and \quad} B(x)=B_0x+B_1x^2+\cdots+B_kx^{k+1}.\]

    [exer:5.5.39] This exercise presents a method for evaluating the integral

    \[y=\int e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)\,dx\]

    where \(\omega\ne0\) and

    \[P(x)=p_0+p_1x+\cdots+p_kx^k,\quad Q(x)=q_0+q_1x+\cdots+q_kx^k.\]

    Show that \(y=e^{\lambda x}u\), where

    \[u'+\lambda u=P(x)\cos \omega x+Q(x)\sin \omega x. \eqno{\rm (A)}\]

    Show that (A) has a particular solution of the form

    \[u_p=A(x)\cos \omega x+B(x)\sin \omega x,\]

    where

    \[A(x)=A_0+A_1x+\cdots+A_kx^k,\quad B(x)=B_0+B_1x+\cdots+B_kx^k,\]

    and the pairs of coefficients \((A_k,B_k)\), \((A_{k-1},B_{k-1})\), …,\((A_0,B_0)\) can be computed successively as the solutions of pairs of equations obtained by equating the coefficients of \(x^r\cos\omega x\) and \(x^r\sin\omega x\) for \(r=k\), \(k-1\), …, \(0\).

    Conclude that

    \[\int e^{\lambda x}\left(P(x)\cos \omega x+Q(x)\sin \omega x\right)\,dx = e^{\lambda x}\left(A(x)\cos \omega x+B(x)\sin \omega x\right) +c,\]

    where \(c\) is a constant of integration.

    [exer:5.5.40] Use the method of Exercise [exer:5.5.39} to evaluate the integral.

    a

    \(\int x^2\cos x\,dx\)

    b

    \(\int x^2e^x\cos x\,dx\)

    [2]

    c

    \(\int xe^{-x}\sin2x\,dx\)

    d

    \(\int x^2e^{-x}\sin x\,dx\)

    [2]

    e

    \(\int x^3e^x\sin x\,dx\)

    f

    \(\int e^x\left[x\cos x-(1+3x)\sin x\right]\,dx\)

    [2]

    g

    \(\int e^{-x}\left[(1+x^2)\cos x+(1-x^2)\sin x\right]\,dx\)