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Mathematics LibreTexts

2.4E: Exercises

  • Page ID
    17372
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    In Exercises \((2.4E.1)\) to \((2.4E.14)\), find a particular solution.

    Exercise \(\PageIndex{1}\)

    \(y''-3y'+2y=e^{3x}(1+x)\)

    Answer

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    Exercise \(\PageIndex{2}\)

    \(y''-6y'+5y=e^{-3x}(35-8x)\)

    Answer

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    Exercise \(\PageIndex{3}\)

    \(y''-2y'-3y=e^x(-8+3x)\)

    Answer

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    Exercise \(\PageIndex{4}\)

    \(y''+2y'+y=e^{2x}(-7-15x+9x^2)\)

    Answer

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    Exercise \(\PageIndex{5}\)

    \(y''+4y=e^{-x}(7-4x+5x^2)\)

    Answer

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    Exercise \(\PageIndex{6}\)

    \(y''-y'-2y=e^x(9+2x-4x^2)\)

    Answer

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    Exercise \(\PageIndex{7}\)

    \(y''-4y'-5y=-6xe^{-x}\)

    Answer

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    Exercise \(\PageIndex{8}\)

    \(y''-3y'+2y=e^x(3-4x)\)

    Answer

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    Exercise \(\PageIndex{9}\)

    \(y''+y'-12y=e^{3x}(-6+7x)\)

    Answer

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    Exercise \(\PageIndex{10}\)

    \(2y''-3y'-2y=e^{2x}(-6+10x)\)

    Answer

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    Exercise \(\PageIndex{11}\)

    \(y''+2y'+y=e^{-x}(2+3x)\)

    Answer

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    Exercise \(\PageIndex{12}\)

    \(y''-2y'+y=e^x(1-6x)\)

    Answer

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    Exercise \(\PageIndex{13}\)

    \(y''-4y'+4y=e^{2x}(1-3x+6x^2)\)

    Answer

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    Exercise \(\PageIndex{14}\)

    \(9y''+6y'+y=e^{-x/3}(2-4x+4x^2)\)

    Answer

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    In Exercises \((2.4E.15)\) to \((2.4E.19)\), find the general solution.

    Exercise \(\PageIndex{15}\)

    \(y''-3y'+2y=e^{3x}(1+x)\)

    Answer

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    Exercise \(\PageIndex{16}\)

    \(y''-6y'+8y=e^x(11-6x)\)

    Answer

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    Exercise \(\PageIndex{17}\)

    \( y''+6y'+9y=e^{2x}(3-5x)\)

    Answer

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    Exercise \(\PageIndex{18}\)

    \(y''+2y'-3y=-16xe^x\)

    Answer

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    Exercise \(\PageIndex{19}\)

    \(y''-2y'+y=e^x(2-12x)\)

    Answer

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    In Exercises \((2.4E.20)\) to \((2.4E.23)\), solve the initial value problem and plot the solution.

    Exercise \(\PageIndex{20}\)

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

    Answer

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    Exercise \(\PageIndex{21}\)

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

    Answer

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    Exercise \(\PageIndex{22}\)

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

    Answer

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    Exercise \(\PageIndex{23}\)

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

    Answer

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    In Exercises \((2.4E.24)\) to \((2.4E.29)\), use the principle of superposition to find a particular solution.

    Exercise \(\PageIndex{24}\)

    \(y''+y'+y=xe^x+e^{-x}(1+2x)\)

    Answer

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    Exercise \(\PageIndex{25}\)

    \(y''-7y'+12y=-e^x(17-42x)-e^{3x}\)

    Answer

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    Exercise \(\PageIndex{26}\)

    \(y''-8y'+16y=6xe^{4x}+2+16x+16x^2\)

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    Exercise \(\PageIndex{27}\)

    \(y''-3y'+2y=-e^{2x}(3+4x)-e^x\)

    Answer

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    Exercise \(\PageIndex{28}\)

    \(y''-2y'+2y=e^x(1+x)+e^{-x}(2-8x+5x^2)\)

    Answer

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    Exercise \(\PageIndex{29}\)

    \(y''+y=e^{-x}(2-4x+2x^2)+e^{3x}(8-12x-10x^2)\)

    Answer

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    Exercise \(\PageIndex{30}\)

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

    \begin{equation}\label{eq:2.4E.1}
    ay''+by'+cy=e^{\alpha x}G(x)
    \end{equation}

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

    \begin{equation}\label{eq:2.4E.2}
    au''+p'(\alpha)u'+p(\alpha)u=G(x)
    \end{equation}

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

    \begin{eqnarray*}
    ay''+by'+cy=0.
    \end{eqnarray*}

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

    \begin{eqnarray*}
    G(x)=g_0+g_1x+g_2x^2+g_3x^3.
    \end{eqnarray*}

    (b) Prove that if \(e^{\alpha x}\) isn't a solution of the complementary equation then \eqref{eq:2.4E.2} 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 \((2.4.4)\). Conclude that \eqref{eq:2.4E.1} has a particular solution of the form \(y_p=e^{\alpha x}A(x)\).

    (c) Show that if \(e^{\alpha x}\) is a solution of the complementary equation and \(xe^{\alpha x}\) isn't, then \eqref{eq:2.4E.2} 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 \((2.4.5)\). Conclude that \eqref{eq:2.4E.1} has a particular solution of the form \(y_p=xe^{\alpha x}A(x)\).

    (d) Show that if \(e^{\alpha x}\) and \(xe^{\alpha x}\) are both solutions of the complementary equation then \eqref{eq:2.4E.2} 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 \((2.4.6)\). Conclude that \eqref{eq:2.4E.1} has a particular solution of the form \(y_p=x^2e^{\alpha x}A(x)\).

    Answer

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    Exercises \((2.4E.31)\) to \((2.4E.36)\) treat the equations considered in Examples \((2.4.1)\) to \((2.4.6)\). Substitute the suggested form of \(y_p\) into the equation and equate the resulting coefficients of like functions on the two sides of the resulting equation to derive a set of simultaneous equations for the coefficients in \(y_p\). Then solve for the coefficients to obtain \(y_p\). Compare the work you've done with the work required to obtain the same results in Examples \((2.4.1)\) to \((2.4.6)\).

    Exercise \(\PageIndex{31}\)

    Compare with Example \((2.4.1)\):

    \begin{eqnarray*}
    y''-7y'+12y=4e^{2x};\quad y_p=Ae^{2x}
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{32}\)

    Compare with Example \((2.4.2)\):

    \begin{eqnarray*}
    y''-7y'+12y=5e^{4x};\quad y_p=Axe^{4x}
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{33}\)

    Compare with Example \((2.4.3)\):

    \begin{eqnarray*}
    y''-8y'+16y=2e^{4x};\quad y_p=Ax^2e^{4x}
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{34}\)

    Compare with Example \((2.4.4)\):

    \begin{eqnarray*}
    y''-3y'+2y=e^{3x}(-1+2x+x^2),\quad y_p=e^{3x}(A+Bx+Cx^2)
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{35}\)

    Compare with Example \((2.4.5)\):

    \begin{eqnarray*}
    y''-4y'+3y=e^{3x}(6+8x+12x^2),\quad y_p=e^{3x}(Ax+Bx^2+Cx^3)
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{36}\)

    Compare with Example \((2.4.6)\):

    \begin{eqnarray*}
    4y''+4y'+y=e^{-x/2}(-8+48x+144x^2),\quad y_p=e^{-x/2}(Ax^2+Bx^3+Cx^4)
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{37}\)

    Write \(y=ue^{\alpha x}\) to find the general solution.

    (a) \(y''+2y'+y=\displaystyle{e^{-x}\over\sqrt x}\)

    (b) \(y''+6y'+9y=e^{-3x}\ln x\)

    (c) \(y''-4y'+4y=\displaystyle{e^{2x}\over1+x}\)

    (d) \(4y''+4y'+y=\displaystyle{4e^{-x/2}\left({1\over x}+x\right)}\)

    Answer

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    Exercise \(\PageIndex{38}\)

    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

    \begin{eqnarray*}
    y=\int e^{\alpha x}P(x)\,dx
    \end{eqnarray*}

    with

    \begin{eqnarray*}
    P(x)=p_0+p_1x+\cdots+p_kx^k, \mbox{\quad (where $p_k\ne0$)}.
    \end{eqnarray*}

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

    \begin{equation}\label{eq:2.4E.3}
    u'+\alpha u=P(x).
    \end{equation}

    (b) Show that \eqref{eq:2.4E.3} has a particular solution of the form

    \begin{eqnarray*}
    u_p=A_0+A_1x+\cdots+A_kx^k,
    \end{eqnarray*}

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

    \begin{eqnarray*}
    u_p'+\alpha u_p=P(x).
    \end{eqnarray*}

    (c) Conclude that

    \begin{eqnarray*}
    \int e^{\alpha x}P(x)\,dx=\left(A_0+A_1x+\cdots+A_kx^k\right)e^{\alpha x} +c,
    \end{eqnarray*}

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

    Answer

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    Exercise \(\PageIndex{39}\)

    Use the method of Exercise \((2.4E.38)\) to evaluate the integral.

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

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

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

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

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

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

    Answer

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    Exercise \(\PageIndex{40}\)

    Use the method suggested in Exercise \((2.4E.38)\) to evaluate \(\int x^ke^{\alpha x}\,dx\), where \(k\) is an arbitrary positive integer and \(\alpha\ne0\).

    Answer

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