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Mathematics LibreTexts

5.4E: The Method of Undetermined Coefficients I (Exercises)

  • Page ID
    18319
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    In Exercises [exer:5.4.1} –[exer:5.4.14} find a particular solution.

    [exer:5.4.1] \(y''-3y'+2y=e^{3x}(1+x)\)

    [exer:5.4.2] \(y''-6y'+5y=e^{-3x}(35-8x)\)

    [exer:5.4.3] \(y''-2y'-3y=e^x(-8+3x)\)

    [exer:5.4.4] \(y''+2y'+y=e^{2x}(-7-15x+9x^2)\)

    [exer:5.4.5] \(y''+4y=e^{-x}(7-4x+5x^2)\)

    [exer:5.4.6] \(y''-y'-2y=e^x(9+2x-4x^2)\)

    [exer:5.4.7] \(y''-4y'-5y=-6xe^{-x}\)

    [exer:5.4.8] \(y''-3y'+2y=e^x(3-4x)\)

    [exer:5.4.9] \(y''+y'-12y=e^{3x}(-6+7x)\)

    [exer:5.4.10] \(2y''-3y'-2y=e^{2x}(-6+10x)\)

    [exer:5.4.11] \(y''+2y'+y=e^{-x}(2+3x)\)

    [exer:5.4.12] \(y''-2y'+y=e^x(1-6x)\)

    [exer:5.4.13] \(y''-4y'+4y=e^{2x}(1-3x+6x^2)\)

    [exer:5.4.14] \(9y''+6y'+y=e^{-x/3}(2-4x+4x^2)\)

    [exer:5.4.15] \(y''-3y'+2y=e^{3x}(1+x)\)

    [exer:5.4.16] \(y''-6y'+8y=e^x(11-6x)\)

    [exer:5.4.17] \(y''+6y'+9y=e^{2x}(3-5x)\)

    [exer:5.4.18] \(y''+2y'-3y=-16xe^x\)

    [exer:5.4.19] \(y''-2y'+y=e^x(2-12x)\)

    [exer:5.4.20] \(y''-4y'-5y=9e^{2x}(1+x), \quad y(0)=0,\quad y'(0)=-10\)

    [exer:5.4.21] \(y''+3y'-4y=e^{2x}(7+6x), \quad y(0)=2,\quad y'(0)=8\)

    [exer:5.4.22] \(y''+4y'+3y=-e^{-x}(2+8x), \quad y(0)=1,\quad y'(0)=2\)

    [exer:5.4.23] \(y''-3y'-10y=7e^{-2x}, \quad y(0)=1,\quad y'(0)=-17\)

    [exer:5.4.24] \(y''+y'+y=xe^x+e^{-x}(1+2x)\)

    [exer:5.4.25] \(y''-7y'+12y=-e^x(17-42x)-e^{3x}\)

    [exer:5.4.26] \(y''-8y'+16y=6xe^{4x}+2+16x+16x^2\)

    [exer:5.4.27] \(y''-3y'+2y=-e^{2x}(3+4x)-e^x\)

    [exer:5.4.28] \(y''-2y'+2y=e^x(1+x)+e^{-x}(2-8x+5x^2)\)

    [exer:5.4.29] \(y''+y=e^{-x}(2-4x+2x^2)+e^{3x}(8-12x-10x^2)\)

    [exer:5.4.30]

    Prove that \(y\) is a solution of the constant coefficient equation

    \[ay''+by'+cy=e^{\alpha x}G(x) \eqno{\rm (A)}\nonumber \]

    if and only if \(y=ue^{\alpha x}\), where \(u\) satisfies

    \[au''+p'(\alpha)u'+p(\alpha)u=G(x) \eqno{\rm (B)}\nonumber \]

    and \(p(r)=ar^2+br+c\) is the characteristic polynomial of the complementary equation

    \[ay''+by'+cy=0.\nonumber \]

    For the rest of this exercise, let \(G\) be a polynomial. Give the requested proofs for the case where

    \[G(x)=g_0+g_1x+g_2x^2+g_3x^3.\nonumber \]

    Prove that if \(e^{\alpha x}\) isn’t a solution of the complementary equation then (B) has a particular solution of the form \(u_p=A(x)\), where \(A\) is a polynomial of the same degree as \(G\), as in Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.4} . Conclude that (A) has a particular solution of the form \(y_p=e^{\alpha x}A(x)\).

    Show that if \(e^{\alpha x}\) is a solution of the complementary equation and \(xe^{\alpha x}\) isn’t , then (B) has a particular solution of the form \(u_p=xA(x)\), where \(A\) is a polynomial of the same degree as \(G\), as in Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.5} . Conclude that (A) has a particular solution of the form \(y_p=xe^{\alpha x}A(x)\).

    Show that if \(e^{\alpha x}\) and \(xe^{\alpha x}\) are both solutions of the complementary equation then (B) has a particular solution of the form \(u_p=x^2A(x)\), where \(A\) is a polynomial of the same degree as \(G\), and \(x^2A(x)\) can be obtained by integrating \(G/a\) twice, taking the constants of integration to be zero, as in Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.6} . Conclude that (A) has a particular solution of the form \(y_p=x^2e^{\alpha x}A(x)\).

    [exer:5.4.31] Compare with Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.1} :

    \[y''-7y'+12y=4e^{2x};\quad y_p=Ae^{2x}\nonumber \]

    [exer:5.4.32] Compare with Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.2} :

    \[y''-7y'+12y=5e^{4x};\quad y_p=Axe^{4x}\nonumber \]

    [exer:5.4.33] Compare with Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.3} .

    \[y''-8y'+16y=2e^{4x};\quad y_p=Ax^2e^{4x}\nonumber \]

    [exer:5.4.34] Compare with Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.4} :

    \[y''-3y'+2y=e^{3x}(-1+2x+x^2),\quad y_p=e^{3x}(A+Bx+Cx^2)\nonumber \]

    [exer:5.4.35] Compare with Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.5} :

    \[y''-4y'+3y=e^{3x}(6+8x+12x^2),\quad y_p=e^{3x}(Ax+Bx^2+Cx^3)\nonumber \]

    [exer:5.4.36] Compare with Example

    Example \(\PageIndex{1}\):

    Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.6} :

    \[4y''+4y'+y=e^{-x/2}(-8+48x+144x^2),\quad y_p=e^{-x/2}(Ax^2+Bx^3+Cx^4)\nonumber \]

    [exer:5.4.37] Write \(y=ue^{\alpha x}\) to find the general solution.

    1. \(y''+2y'+y={e^{-x}\over\sqrt x}\)
    2. \(y''+6y'+9y=e^{-3x}\ln x\)
    3. \(y''-4y'+4y={e^{2x}\over1+x}\)
    4. \(4y''+4y'+y={4e^{-x/2}\left({1\over x}+x\right)}\)

    [exer:5.4.38] Suppose \(\alpha\ne0\) and \(k\) is a positive integer. In most calculus books integrals like \(\int x^k e^{\alpha x}\,dx\) are evaluated by integrating by parts \(k\) times. This exercise presents another method. Let

    \[y=\int e^{\alpha x}P(x)\,dx\nonumber \]

    with

    \[P(x)=p_0+p_1x+\cdots+p_kx^k\nonumber \]

    (where \(p_k \neq 0\)).

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

    \[u'+\alpha u=P(x). \eqno{\rm (A)}\nonumber \]

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

    \[u_p=A_0+A_1x+\cdots+A_kx^k,\nonumber \]

    where \(A_k\), \(A_{k-1}\), …, \(A_0\) can be computed successively by equating coefficients of \(x^k,x^{k-1}, \dots,1\) on both sides of the equation

    \[u_p'+\alpha u_p=P(x).\nonumber \]

    Conclude that

    \[\int e^{\alpha x}P(x)\,dx=\left(A_0+A_1x+\cdots+A_kx^k\right)e^{\alpha x} +c,\nonumber \]

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

    [exer:5.4.39] Use the method of Exercise [exer:5.4.38} to evaluate the integral.

    a

    \(\int e^x(4+x)\,dx\)

    b

    \(\int e^{-x}(-1+x^2)\,dx\)

    [10pt]

    c

    \(\int x^3e^{-2x}\,dx\)

    d

    \(\int e^x(1+x)^2\,dx\)

    [10pt]

    e

    \(\int e^{3x}(-14+30x+27x^2)\,dx\)

    f

    \(\int e^{-x}(1+6x^2-14x^3+3x^4)\,dx\)

    [exer:5.4.40] Use the method suggested in Exercise [exer:5.4.38} to evaluate \(\int x^ke^{\alpha x}\,dx\), where \(k\) is an arbitrary positive integer and \(\alpha\ne0\).