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Mathematics LibreTexts

5.2E: Constant Coefficient Homogeneous Equations (Exercises)

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    18317
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    In Exercises [exer:5.2.1} –[exer:5.2.12} find the general solution.

    [exer:5.2.1] \(y''+5y'-6y=0\)

    [exer:5.2.2] \(y''-4y'+5y=0\)

    [exer:5.2.3] \(y''+8y'+7y=0\) [exer:5.2.4] \(y''-4y'+4y=0\)
    [exer:5.2.5] \(y''+2y'+10y=0\) [exer:5.2.6] \(y''+6y'+10y=0\)
    [exer:5.2.7] \(y''-8y'+16y=0\) [exer:5.2.8] \(y''+y'=0\)
    [exer:5.2.9] \(y''-2y'+3y=0\) [exer:5.2.10] \(y''+6y'+13y=0\)
    [exer:5.2.11] \(4y''+4y'+10y=0\) [exer:5.2.12] \(10y''-3y'-y=0\)

    [exer:5.2.13] \(y''+14y'+50y=0, \quad y(0)=2,\quad y'(0)=-17\)

    [exer:5.2.14] \(6y''-y'-y=0, \quad y(0)=10,\quad y'(0)=0\)

    [exer:5.2.15] \(6y''+y'-y=0, \quad y(0)=-1,\quad y'(0)=3\)

    [exer:5.2.16] \(4y''-4y'-3y=0, \quad y(0)={13\over 12},\quad y'(0)={23 \over 24}\)

    [exer:5.2.17] \(4y''-12y'+9y=0, \quad y(0)=3,\quad y'(0)={5\over 2}\)

    [exer:5.2.18] \(y''+7y'+12y=0, \quad y(0)=-1,\quad y'(0)=0\)

    [exer:5.2.19] \(y''-6y'+9y=0, \quad y(0)=0,\quad y'(0)=2\)

    [exer:5.2.20] \(36y''-12y'+y=0, \quad y(0)=3,\quad y'(0)={5\over2}\)

    [exer:5.2.21] \(y''+4y'+10y=0, \quad y(0)=3,\quad y'(0)=-2\)

    [exer:5.2.22]

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

    \[ay''+by'+cy=0. \eqno{\rm (A)}\nonumber \]

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

    \[az''+bz'+cz=0.\nonumber \]

    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 (A). Show that \(\{z_1,z_2\}\) is also a fundamental set of solutions of (A).

    The statement of Theorem [thmtype:5.2.1} is convenient for solving an initial value problem

    \[ay''+by'+cy=0, \quad y(0)=k_0,\quad y'(0)=k_1,\nonumber \]

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

    \[ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1, \eqno{\rm (B)}\nonumber \]

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

    \[y=c_1e^{r_1x}+c_2e^{r_2x}, \quad y=e^{r_1x}(c_1+c_2x),\mbox{ or } y=e^{\lambda x}(c_1\cos\omega x+c_2\sin\omega x)\nonumber \]

    (whichever is applicable) is more complicated. Use

    b

    to restate Theorem [thmtype:5.2.1} in a form more convenient for solving (B).

    [exer:5.2.23] \(y''+3y'+2y=0, \quad y(1)=-1,\quad y'(1)=4\)

    [exer:5.2.24] \(y''-6y'-7y=0, \quad y(2)=-{1\over3},\quad y'(2)=-5\)

    [exer:5.2.25] \(y''-14y'+49y=0, \quad y(1)=2,\quad y'(1)=11\)

    [exer:5.2.26] \(9y''+6y'+y=0, \quad y(2)=2,\quad y'(2)=-{14\over3}\)

    [exer:5.2.27] \(9y''+4y=0, \quad y(\pi/4)=2,\quad y'(\pi/4)=-2\)

    [exer:5.2.28] \(y''+3y=0, \quad y(\pi/3)=2,\quad y'(\pi/3)=-1\)

    [exer:5.2.29] Prove: If the characteristic equation of

    \[ay''+by'+cy=0 \eqno{\rm (A)}\nonumber \]

    has a repeated negative root or two roots with negative real parts, then every solution of (A) approaches zero as \(x\to\infty\).

    [exer:5.2.30] Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has distinct real roots \(r_1\) and \(r_2\). Use a method suggested by Exercise [exer:5.2.22} to find a formula for the solution of

    \[ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.\nonumber \]

    [exer:5.2.31] Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has a repeated real root \(r_1\). Use a method suggested by Exercise [exer:5.2.22} to find a formula for the solution of

    \[ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.\nonumber \]

    [exer:5.2.32] Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has complex conjugate roots \(\lambda\pm i\omega\). Use a method suggested by Exercise [exer:5.2.22} to find a formula for the solution of

    \[ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.\nonumber \]

    [exer:5.2.33] Suppose the characteristic equation of

    \[ay''+by'+cy=0 \eqno{\rm (A)}\nonumber \]

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

    Show that

    \[a{\partial^2\over\partial^2 x}(e^{rx})+b{\partial \over\partial x}(e^{rx}) +ce^{rx}=a(r-r_1)^2e^{rx}. \eqno{\rm (B)}\nonumber \]

    Differentiate (B) with respect to \(r\) to obtain

    \[a{\partial\over\partial r}\left({\partial^2\over\partial^2 x}(e^{rx})\right)+b{\partial\over\partial r}\left({\partial \over\partial x}(e^{rx})\right) +c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \eqno{\rm (C)}\nonumber \]

    Reverse the orders of the partial differentiations in the first two terms on the left side of (C) to obtain

    \[a{\partial^2\over\partial x^2}(xe^{rx})+b{\partial\over\partial x}(xe^{rx})+c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \eqno{\rm (D)}\nonumber \]

    Set \(r=r_1\) in (B) and (D) to see that \(y_1=e^{r_1x}\) and \(y_2=xe^{r_1x}\) are solutions of (A)

    [exer:5.2.34] In calculus you learned that \(e^u\), \(\cos u\), and \(\sin u\) can be represented by the infinite series

    \[e^u=\sum_{n=0}^\infty {u^n\over n!} =1+{u\over 1!}+{u^2\over 2!}+{u^3\over 3!}+\cdots+{u^n\over n!}+\cdots \eqno{\rm (A)}\nonumber \]

    \[\cos u=\sum_{n=0}^\infty (-1)^n{u^{2n}\over(2n)!} =1-{u^2\over2!}+{u^4\over4!}+\cdots+(-1)^n{u^{2n}\over(2n)!} +\cdots, \eqno{\rm (B)}\nonumber \]

    and

    \[\sin u=\sum_{n=0}^\infty (-1)^n{u^{2n+1}\over(2n+1)!} =u-{u^3\over3!}+{u^5\over5!}+\cdots+(-1)^n {u^{2n+1}\over(2n+1)!} +\cdots \eqno{\rm (C)}\nonumber \]

    for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i\theta\), where \(\theta\) is real, to obtain

    \[e^{i\theta}=\sum_{n=0}^\infty {(i\theta)^n\over n!}. \eqno{\rm (D)}\nonumber \]

    Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\).

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

    \[1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,\cdots.\nonumber \]

    Use this to group the terms in (D) as

    \[\begin{aligned} e^{i\theta}&=&\left(1-{\theta^2\over2}+{\theta^4\over4}+\cdots\right) +i\left(\theta-{\theta^3\over3!}+{\theta^5\over5!}+\cdots\right)\

    \[2\jot] &=&\sum_{n=0}^\infty (-1)^n{\theta^{2n}\over(2n)!} +i\sum_{n=0}^\infty (-1)^n{\theta^{2n+1}\over(2n+1)!}.\end{aligned}\nonumber \]

    By comparing this result with (B) and (C), conclude that

    \[e^{i\theta}=\cos\theta+i\sin\theta. \eqno{\rm (E)}\nonumber \]

    This is Euler’s identity.

    Starting from

    \[e^{i\theta_1}e^{i\theta_2}=(\cos\theta_1+i\sin\theta_1) (\cos\theta_2+i\sin\theta_2),\nonumber \]

    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

    \[\begin{aligned} \cos(\theta_1+\theta_2)&=&\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\\ \sin(\theta_1+\theta_2)&=&\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\end{aligned}\nonumber \]

    to verify that

    \[e^{i(\theta_1+\theta_2)}=e^{i\theta_1}e^{i\theta_2},\nonumber \]

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

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

    \[e^{\alpha+i\beta}=e^\alpha e^{i\beta}=e^\alpha(\cos\beta+i\sin\beta). \eqno{\rm (F)}\nonumber \]

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

    \[e^{z_1+z_2}=e^{z_1}e^{z_2}.\nonumber \]

    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

    \[ay''+by'+cy=0 \eqno{\rm (G)}\nonumber \]

    if \(u\) and \(v\) are both solutions of (G). Use Theorem [thmtype:5.2.1}

    c

    to verify that if the characteristic equation of (G) 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 (G).