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Mathematics LibreTexts

2.2E: Exercises

  • Page ID
    17354
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    In Exercises \((2.2E.1)\) to \((2.2E.12)\), find the general solution.

    Exercise \(\PageIndex{1}\)

    \(y''+5y'-6y=0\)

     

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    Exercise \(\PageIndex{2}\)

    \(y''-4y'+5y=0\)

     

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    Exercise \(\PageIndex{3}\)

    \(y''+8y'+7y=0\)

     

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    Exercise \(\PageIndex{4}\)

    \(y''-4y'+4y=0\)

     

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    Exercise \(\PageIndex{5}\)

    \(y''+2y'+10y=0\)

     

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    Exercise \(\PageIndex{6}\)

    \(y''+6y'+10y=0\)

     

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    Exercise \(\PageIndex{7}\)

    \(y''-8y'+16y=0\)

     

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    Exercise \(\PageIndex{8}\)

    \(y''+y'=0\)

     

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    Exercise \(\PageIndex{9}\)

    \(y''-2y'+3y=0\)

     

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    Exercise \(\PageIndex{10}\)

    \(y''+6y'+13y=0 \)

     

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    Exercise \(\PageIndex{11}\)

    \(4y''+4y'+10y=0\)

     

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    Exercise \(\PageIndex{12}\)

    \(10y''-3y'-y=0\)

     

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    In Exercises \((2.2E.13)\) to \((2.2E.17)\), solve the initial value problem.

    Exercise \(\PageIndex{13}\)

    \(y''+14y'+50y=0, \quad y(0)=2,\quad y'(0)=-17\)

     

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    Exercise \(\PageIndex{14}\)

    \(6y''-y'-y=0, \quad y(0)=10,\quad y'(0)=0\)

     

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    Exercise \(\PageIndex{15}\)

    \(6y''+y'-y=0, \quad y(0)=-1,\quad y'(0)=3\)

     

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    Exercise \(\PageIndex{16}\)

    \(4y''-4y'-3y=0, \quad y(0)=\displaystyle{13\over 12},\quad y'(0)=\displaystyle{23 \over 24}\)

     

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    Exercise \(\PageIndex{17}\)

    \(4y''-12y'+9y=0, \quad y(0)=3,\quad y'(0)=\displaystyle{5\over 2}\)

     

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    In Exercises \9(2.2E.18)\) to \((2.2E.21)\), solve the initial value problem and graph the solution.

    Exercise \(\PageIndex{18}\)

    \(y''+7y'+12y=0, \quad y(0)=-1,\quad y'(0)=0\)

     

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    Exercise \(\PageIndex{19}\)

    \(y''-6y'+9y=0, \quad y(0)=0,\quad y'(0)=2\)

     

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    Exercise \(\PageIndex{20}\)

    \(36y''-12y'+y=0, \quad y(0)=3,\quad y'(0)=\displaystyle{5\over2}\)

     

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    Exercise \(\PageIndex{21}\)

    \(y''+4y'+10y=0, \quad y(0)=3,\quad y'(0)=-2\)

     

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    Exercise \(\PageIndex{22}\)

    (a) Suppose \(y\) is a solution of the constant coefficient homogeneous equation

    \begin{equation}\label{eq:2.2E.1}
    ay''+by'+cy=0.
    \end{equation}

    Let \(z(x)=y(x-x_0)\), where \(x_0\) is an arbitrary real number. Show that

    \begin{eqnarray*}
    az''+bz'+cz=0.
    \end{eqnarray*}

     

    (b) Let \(z_1(x)=y_1(x-x_0)\) and \(z_2(x)=y_2(x-x_0)\), where \(\{y_1,y_2\}\) is a fundamental set of solutions of \eqref{eq:2.2E.1} (A). Show that \(\{z_1,z_2\}\) is also a fundamental set of solutions of \eqref{eq:2.2E.1}.

     

    (c) The statement of Theorem \((2.2.1)\) is convenient for solving an initial value problem

    \begin{eqnarray*}
    ay''+by'+cy=0, \quad y(0)=k_0,\quad y'(0)=k_1,
    \end{eqnarray*}

    where the initial conditions are imposed at \(x_0=0\). However, if the initial value problem is

    \begin{equation}\label{eq:2.2E.2}
    ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1,
    \end{equation}

    where \(x_0\ne0\), then determining the constants in

    \begin{eqnarray*}
    y=c_1e^{r_1x}+c_2e^{r_2x}, \quad y=e^{r_1x}(c_1+c_2x),\mbox{ or } y=e^{\lambda x}(c_1\cos\omega x+c_2\sin\omega x)
    \end{eqnarray*}

    (whichever is applicable) is more complicated. Use part (b) to restate Theorem \((2.2.1)\) in a form more convenient for solving \eqref{eq:2.2E.2}.

     

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    In Exercises \((2.2E.23)\) tp \((2.2E.28)\), use a method suggested by Exercise \((2.2E.22)\) to solve the initial value problem.

    Exercise \(\PageIndex{23}\)

    \(y''+3y'+2y=0, \quad y(1)=-1,\quad y'(1)=4\)

     

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    Exercise \(\PageIndex{24}\)

    \(y''-6y'-7y=0, \quad y(2)=-\displaystyle{1\over3},\quad y'(2)=-5\)

     

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    Exercise \(\PageIndex{25}\)

    \(y''-14y'+49y=0, \quad y(1)=2,\quad y'(1)=11\)

     

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    Exercise \(\PageIndex{26}\)

    \(9y''+6y'+y=0, \quad y(2)=2,\quad y'(2)=-\displaystyle{14\over3}\)

     

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    Exercise \(\PageIndex{27}\)

    \(9y''+4y=0, \quad y(\pi/4)=2,\quad y'(\pi/4)=-2\)

     

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    Exercise \(\PageIndex{28}\)

    \(y''+3y=0, \quad y(\pi/3)=2,\quad y'(\pi/3)=-1\)

     

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    Exercise \(\PageIndex{29}\)

    Prove: If the characteristic equation of

    \begin{equation}\label{eq:2.2E.3}
    ay''+by'+cy=0
    \end{equation}

    has a repeated negative root or two roots with negative real parts, then every solution of \eqref{eq:2.2E.3} approaches zero as \(x\to\infty\).

     

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    Exercise \(\PageIndex{30}\)

    Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has distinct real roots \(r_1\) and \(r_2\). Use a method suggested by Exercise \((2.2E.22)\) to find a formula for the solution of

    \begin{eqnarray*}
    ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.
    \end{eqnarray*}

     

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    Exercise \(\PageIndex{31}\)

    Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has a repeated real root \(r_1\). Use a method suggested by Exercise \((2.2E.22)\) to find a formula for the solution of 

    \begin{eqnarray*}
    ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.
    \end{eqnarray*}

     

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    Exercise \(\PageIndex{32}\)

    Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has complex conjugate roots \(\lambda\pm i\omega\). Use a method suggested by Exercise \((2.2E.22)\) to find a formula for the solution of

    \begin{eqnarray*}
    ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.
    \end{eqnarray*}

     

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    Exercise \(\PageIndex{33}\)

    Suppose the characteristic equation of

    \begin{equation}\label{eq:2.2E.4}
    ay''+by'+cy=0
    \end{equation}

    has a repeated real root \(r_1\). Temporarily, think of \(e^{rx}\) as a function of two real variables \(x\) and \(r\).

    (a) Show that

    \begin{equation}\label{eq:2.2E.5}
    a{\partial^2\over\partial^2 x}(e^{rx})+b{\partial \over\partial x}(e^{rx}) +ce^{rx}=a(r-r_1)^2e^{rx}.
    \end{equation}

     

    (b) Differentiate \eqref{eq:2.2E.5} with respect to \(r\) to obtain

    \begin{equation}\label{eq:2.2E.6}
    a{\partial\over\partial r}\left({\partial^2\over\partial^2x}(e^{rx})\right)+b{\partial\over\partial r}\left({\partial \over\partial x}(e^{rx})\right) +c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}.
    \end{equation}

     

    (c) Reverse the orders of the partial differentiations in the first two terms on the left side of \eqref{eq:2.2E.6} to obtain

    \begin{equation}\label{eq:2.2E.7}
    a{\partial^2\over\partial x^2}(xe^{rx})+b{\partial\over\partial x}(xe^{rx})+c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}.
    \end{equation}

     

    (d) Set \(r=r_1\) in \eqref{eq:2.2E.5} and \eqref{eq:2.2E.7} to see that \(y_1=e^{r_1x}\) and \(y_2=xe^{r_1x}\) are solutions of \eqref{eq:2.2E.4}.

     

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    Exercise \(\PageIndex{34}\)

    In calculus you learned that \(e^u\), \(\cos u\), and \(\sin u\) can be represented by the infinite series

    \begin{equation}\label{eq:2.2E.8}
    e^u=\sum_{n=0}^\infty {u^n\over n!} =1+{u\over 1!}+{u^2\over 2!}+{u^3\over 3!}+\cdots+{u^n\over n!}+\cdots
    \end{equation}

    \begin{equation}\label{eq:2.2E.9}
    \cos u=\sum_{n=0}^\infty (-1)^n{u^{2n}\over(2n)!} =1-{u^2\over2!}+{u^4\over4!}+\cdots+(-1)^n{u^{2n}\over(2n)!} +\cdots,
    \end{equation}

    and

    \begin{equation}\label{eq:2.2E.10}
    \sin u=\sum_{n=0}^\infty (-1)^n{u^{2n+1}\over(2n+1)!} =u-{u^3\over3!}+{u^5\over5!}+\cdots+(-1)^n {u^{2n+1}\over(2n+1)!} +\cdots
    \end{equation}

    for all real values of \(u\). Even though you have previously considered \eqref{eq:2.2E.8} only for real values of \(u\), we can set \(u=i\theta\), where \(\theta\) is real, to obtain

    \begin{equation}\label{eq:2.2E.11}
    e^{i\theta}=\sum_{n=0}^\infty {(i\theta)^n\over n!}.
    \end{equation}

    Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in \eqref{eq:2.2E.11} converges for all real \(\theta\). 

    (a) Recalling that \(i^2=-1,\) write enough terms of the sequence \(\{i^n\}\) to convince yourself that the sequence is repetitive:

    \begin{eqnarray*}
    1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,\cdots.
    \end{eqnarray*}

    Use this to group the terms in \eqref{eq:2.2E.11} as

    \begin{eqnarray*}
    e^{i\theta}&=&\left(1-{\theta^2\over2}+{\theta^4\over4}+\cdots\right) +i\left(\theta-{\theta^3\over3!}+{\theta^5\over5!}+\cdots\right)\\
    &=&\sum_{n=0}^\infty (-1)^n{\theta^{2n}\over(2n)!} +i\sum_{n=0}^\infty (-1)^n{\theta^{2n+1}\over(2n+1)!}.
    \end{eqnarray*}

    By comparing this result with \eqref{eq:2.2E.9} and \eqref{eq:2.2E.10}, conclude that 

    \begin{equation}\label{eq:2.2E.12}
    e^{i\theta}=\cos\theta+i\sin\theta.
    \end{equation}

    This is Euler's Identity

     

    (b) Starting from

    \begin{eqnarray*}
    e^{i\theta_1}e^{i\theta_2}=(\cos\theta_1+i\sin\theta_1), (\cos\theta_2+i\sin\theta_2),
    \end{eqnarray*}

    collect the real part (the terms not multiplied by \(i\)) and the imaginary part (the terms multiplied by \(i\)) on the right, and use the trigonometric identities

    \begin{eqnarray*}
    \cos(\theta_1+\theta_2)&=&\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\\
    \sin(\theta_1+\theta_2)&=&\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2
    \end{eqnarray*}

    to verify that

    \begin{eqnarray*}
    e^{i(\theta_1+\theta_2)}=e^{i\theta_1}e^{i\theta_2},
    \end{eqnarray*}

    as you would expect from the use of the exponential notation \(e^{i\theta}\).

     

    (c) If \(\alpha\) and \(\beta\) are real numbers, define

    \begin{equation}\label{eq:2.2E.13}
    e^{\alpha+i\beta}=e^\alpha e^{i\beta}=e^\alpha(\cos\beta+i\sin\beta).
    \end{equation}

    Show that if \(z_1=\alpha_1+i\beta_1\) and \(z_2=\alpha_2+i\beta_2\) then

    \begin{eqnarray*}
    e^{z_1+z_2}=e^{z_1}e^{z_2}.
    \end{eqnarray*}

     

    (d) Let \(a\), \(b\), and \(c\) be real numbers, with \(a\ne0\). Let \(z=u+iv\) where \(u\) and \(v\) are real-valued functions of \(x\). Then we say that \(z\) is a solution of

    \begin{equation}\label{eq:2.2E.14}
    ay''+by'+cy=0
    \end{equation}

    if \(u\) and \(v\) are both solutions of \eqref{eq:2.2E.14}. Use Theorem \((2.2.1)\) (c) to verify that if the characteristic equation of \eqref{eq:2.2E.14} has complex conjugate roots \(\lambda\pm i\omega\) then \(z_1=e^{(\lambda+i\omega)x}\) and \(z_2=e^{(\lambda-i\omega)x}\) are both solutions of \eqref{eq:2.2E.14}.

     

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