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Mathematics LibreTexts

9.2E: Higher Order Constant Coefficient Homogeneous Equations (Exercises)

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    In Exercises [exer:9.2.1}– [exer:9.2.14} find the general solution.

    [exer:9.2.1] \(y'''-3y''+3y'-y=0\)

    [exer:9.2.2] \(y^{(4)}+8y''-9y=0\)

    [exer:9.2.3] \(y'''-y''+16y'-16y=0\)

    [exer:9.2.4] \(2y'''+3y''-2y'-3y=0\)

    [exer:9.2.5] \(y'''+5y''+9y'+5y=0\)

    [exer:9.2.6] \(4y'''-8y''+5y'-y=0\)

    [exer:9.2.7] \(27y'''+27y''+9y'+y=0\)

    [exer:9.2.8] \(y^{(4)}+y''=0\)

    [exer:9.2.9] \(y^{(4)}-16y=0\)

    [exer:9.2.10] \(y^{(4)}+12y''+36y=0\)

    [exer:9.2.11] \(16y^{(4)}-72y''+81y=0\)

    [exer:9.2.12] \(6y^{(4)}+5y'''+7y''+5y'+y=0\)

    [exer:9.2.13] \(4y^{(4)}+12y'''+3y''-13y'-6y=0\)

    [exer:9.2.14] \(y^{(4)}-4y'''+7y''-6y'+2y=0\)

    [exer:9.2.15] \(y'''-2y''+4y'-8y=0, \quad y(0)=2,\quad y'(0)=-2,\; y''(0)=0\)

    [exer:9.2.16] \(y'''+3y''-y'-3y=0, \quad y(0)=0,\quad y'(0)=14,\quad y''(0)=-40\)

    [exer:9.2.17] \(y'''-y''-y'+y=0, \quad y(0)=-2,\quad y'(0)=9,\quad y''(0)=4\)

    [exer:9.2.18] \(y'''-2y'-4y=0, \quad y(0)=6,\quad y'(0)=3,\quad y''(0)=22\)

    [exer:9.2.19] \(3y'''-y''-7y'+5y=0, \quad y(0)= 14\over5,\quad y'(0)=0,\quad y''(0)=10\)

    [exer:9.2.20] \(y'''-6y''+12y'-8y=0, \quad y(0)=1,\quad y'(0)=-1,\quad y''(0)=-4\)

    [exer:9.2.21] \(2y'''-11y''+12y'+9y=0, \quad y(0)=6,\quad y'(0)=3,\quad y''(0)=13\)

    [exer:9.2.22] \(8y'''-4y''-2y'+y=0, \quad y(0)=4,\quad y'(0)=-3,\quad y''(0)=-1\)

    [exer:9.2.23] \(y^{(4)}-16y=0, \quad y(0)=2,\; y'(0)=2,\; y''(0)=-2,\; y'''(0)=0\)

    [exer:9.2.24] \(y^{(4)}-6y'''+7y''+6y'-8y=0, \quad y(0)=-2,\quad y'(0)=-8,\quad y''(0)=-14\),


    [exer:9.2.25] \(4y^{(4)}-13y''+9y=0, \quad y(0)=1,\quad y'(0)=3,\quad y''(0)=1,\quad y'''(0)=3\)

    [exer:9.2.26] \(y^{(4)}+2y'''-2y''-8y'-8y=0, \quad y(0)=5,\quad y'(0)=-2,\quad y''(0)=6,\quad y'''(0)=8\)

    [exer:9.2.27] \(4y^{(4)}+8y'''+19y''+32y'+12y=0, \quad y(0)=3,\quad y'(0)=-3,\quad y''(0)= -7\over2\), \(y'''(0)=31\over4}\)

    [exer:9.2.28] Find a fundamental set of solutions of the given equation, and verify that it is a fundamental set by evaluating its Wronskian at \(x=0\).

    1. \((D-1)^2(D-2)y=0\)
    2. \((D^2+4)(D-3)y=0\)
    3. \((D^2+2D+2)(D-1)y=0\)
    4. \(D^3(D-1)y=0\)
    5. \((D^2-1)(D^2+1)y=0\)
    6. \((D^2-2D+2)(D^2+1)y=0\)

    [exer:9.2.29] \((D^2+6D+13)(D-2)^2D^3y=0\)

    [exer:9.2.30] \((D-1)^2(2D-1)^3(D^2+1)y=0\)

    [exer:9.2.31] \((D^2+9)^3D^2y=0\) [exer:9.2.32] \((D-2)^3(D+1)^2Dy=0\) [exer:9.2.33] \((D^2+1)(D^2+9)^2(D-2)y=0\) [exer:9.2.34] \((D^4-16)^2y=0\)

    [exer:9.2.35] \((4D^2+4D+9)^3y=0\)

    [exer:9.2.36] \(D^3(D-2)^2(D^2+4)^2y=0\)

    [exer:9.2.37] \((4D^2+1)^2(9D^2+4)^3y=0\) [exer:9.2.38] \(\left[(D-1)^4-16\right]y=0\)

    [exer:9.2.39] It can be shown that \[\left|\begin{array}{cccc} 1&1&\cdots&1\\[4pt] a_1&a_2&\cdots&a_n\\[4pt] a^2_1&a^2_2&\cdots&a^2_n\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] a^{n-1}_1&a^{n-1}_2&\cdots&a^{n-1}_n\end{array}\right|= \prod_{1\le i

    where the left side is the Vandermonde determinant and the right side is the product of all factors of the form \((a_j-a_i)\) with \(i\) and \(j\) between \(1\) and \(n\) and \(i<j\).>

    Verify (A) for \(n=2\) and \(n=3\).

    Find the Wronskian of \(\{e^{{a_1}x}, \quad e^{{a_2}x},\dots, e^{{a_n}x}\}\).

    [exer:9.2.40] A theorem from algebra says that if \(P_1\) and \(P_2\) are polynomials with no common factors then there are polynomials \(Q_1\) and \(Q_2\) such that \[Q_1P_1+Q_2P_2=1.\nonumber\] This implies that \[Q_1(D)P_1(D)y+Q_2(D)P_2(D)y=y\nonumber\] for every function \(y\) with enough derivatives for the left side to be defined.

    Use this to show that if \(P_1\) and \(P_2\) have no common factors and \[P_1(D)y=P_2(D)y=0\nonumber\] then \(y=0\).

    Suppose \(P_1\) and \(P_2\) are polynomials with no common factors. Let \(u_1\), …, \(u_r\) be linearly independent solutions of \(P_1(D)y=0\) and let \(v_1\), …, \(v_s\) be linearly independent solutions of \(P_2(D)y=0\). Use (a) to show that \(\{u_1,\dots,u_r,\allowbreak v_1,\dots,v_s\}\) is a linearly independent set.

    Suppose the characteristic polynomial of the constant coefficient equation \[a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=0 \tag{A}\] has the factorization \[p(r)=a_0p_1(r)p_2(r)\cdots p_k(r),\nonumber\] where each \(p_j\) is of the form \[p_j(r)=(r-r_j)^{n_j} \mbox{ or } p_j(r)=[(r-\lambda_j)^2+w^2_j]^{m_j}\quad (\omega_j>0)\nonumber\] and no two of the polynomials \(p_1\), \(p_2\), …, \(p_k\) have a common factor. Show that we can find a fundamental set of solutions \(\{y_1,y_2,\dots,y_n\}\) of (A) by finding a fundamental set of solutions of each of the equations \[p_j(D)y=0,\quad 1\le j\le k,\nonumber\] and taking \(\{y_1,y_2,\dots,y_n\}\) to be the set of all functions in these separate fundamental sets.

    [exer:9.2.41] Show that if \[z=p(x)\cos\omega x+q(x)\sin\omega x, \tag{A}\] where \(p\) and \(q\) are polynomials of degree \(\le k\), then \[(D^2+\omega^2)z=p_1(x)\cos\omega x+q_1(x)\sin\omega x,\nonumber\] where \(p_1\) and \(q_1\) are polynomials of degree \(\le k-1\).

    Apply (a) \(m\) times to show that if \(z\) is of the form (A) where \(p\) and \(q\) are polynomial of degree \(\le m-1\), then \[(D^2+\omega^2)^mz=0. \tag{B}\]

    Use Equation \ref{eq:9.2.17} to show that if \(y=e^{\lambda x}z\) then \[[(D-\lambda)^2+\omega^2]^my=e^{\lambda x}(D^2+\omega^2)^mz.\nonumber\] Conclude from (b) and (c) that if \(p\) and \(q\) are arbitrary polynomials of degree \(\le m-1\) then \[y=e^{\lambda x}(p(x)\cos\omega x+q(x)\sin\omega x)\nonumber\] is a solution of \[[(D-\lambda)^2+\omega^2]^my=0. \tag{C}\] Conclude from (d) that the functions \[\begin{array}{rl} e^{\lambda x}\cos\omega x, xe^{\lambda x}\cos\omega x, &\dots, x^{m-1}e^{\lambda x}\cos\omega x,\\ e^{\lambda x}\sin\omega x, xe^{\lambda x}\sin\omega x,& \dots, x^{m-1}e^{\lambda x}\sin\omega x \end{array} \tag{\rm (D)}\nonumber\] are all solutions of (C).

    Complete the proof of Theorem [thmtype:9.2.2} by showing that the functions in (D) are linearly independent.

    [exer:9.2.42] Use the trigonometric identities \[\begin{aligned} \cos(A+B)&=&\cos A\cos B-\sin A\sin B\\ \sin(A+B)&=&\cos A\sin B+\sin A\cos B\end{aligned}\nonumber\] to show that \[(\cos A+i\sin A)(\cos B+i\sin B)=\cos(A+B)+i\sin(A+B).\nonumber\]

    Apply (a) repeatedly to show that if \(n\) is a positive integer then \[\prod_{k=1}^n(\cos A_k+i\sin A_k)=\cos(A_1+A_2+\cdots+A_n) +i\sin(A_1+A_2+\cdots+A_n).\nonumber\]

    Infer from (b) that if \(n\) is a positive integer then \[(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta. \tag{A}\]

    Show that (A) also holds if \(n=0\) or a negative integer.

    Now suppose \(n\) is a positive integer. Infer from (A) that if \[z_k=\cos\left(2k\pi\over n\right)+i\sin\left(2k\pi\over n\right) ,\quad k=0,1,\dots,n-1,\nonumber\] and \[\zeta_k=\cos\left((2k+1)\pi\over n\right)+i\sin\left((2k+1)\pi\over n\right) ,\quad k=0,1,\dots,n-1,\nonumber\] then \[z_k^n=1\quad\mbox{ and }\quad\zeta_k^n=-1,\quad k=0,1,\dots,n-1.\nonumber\]

    (Why don’t we also consider other integer values for \(k\)?)

    Let \(\rho\) be a positive number. Use (e) to show that \[z^n-\rho=(z-\rho^{1/n} z_0)(z-\rho^{1/n}z_1)\cdots(z-\rho^{1/n} z_{n-1})\nonumber\] and \[z^n+\rho=(z-\rho^{1/n} \zeta_0)(z-\rho^{1/n} \zeta_1)\cdots(z-\rho^{1/n} \zeta_{n-1}).\nonumber\]

    [exer:9.2.43] Use (e) of Exercise [exer:9.2.42} to find a fundamental set of solutions of the given equation.

    1. \(y'''-y=0\)
    2. \(y'''+y=0\)
    3. \(y^{(4)}+64y=0\)
    4. \(y^{(6)}-y=0\)
    5. \(y^{(6)}+64y=0\)
    6. \(\left[(D-1)^6-1\right]y=0\)
    7. \(y^{(5)}+y^{(4)}+y'''+y''+y'+y=0\)

    [exer:9.2.44] An equation of the form \[a_0x^ny^{(n)}+a_1x^{n-1}y^{(n-1)}+\cdots +a_{n-1}xy'+a_ny=0,\quad x>0, \tag{A}\] where \(a_0\), \(a_1\), …, \(a_n\) are constants, is an Euler or equidimensional equation. Show that if \[x=e^t \quad \mbox{ and } \quad Y(t)=y(x(t)), \tag{B}\] then \[\begin{aligned} \ x {dy\over dx}&=&\{dY\over dt}\\[4pt] \ x^2{d^2y\over dx^2}&=&\{d^2Y\over dt^2}-{dY\over dt}\\[4pt] \ x^3{d^3y\over dx^3}&=&\{d^3Y\over dt^3}-3{d^2Y\over dt^2}+2{dY\over dt}.\end{aligned}\nonumber\]

    In general, it can be shown that if \(r\) is any integer \(\ge2\) then

    \[x^r {d^ry\over dx^r}={d^rY\over dt^r}+ A_{1r}{d^{r-1}Y\over dt^{r-1}}+\cdots+A_{r-1,r} {dY\over dt}\nonumber\]

    where \(A_{1r}\), …, \(A_{r-1,r}\) are integers. Use these results to show that the substitution (B) transforms (A) into a constant coefficient equation for \(Y\) as a function of \(t\).

    [exer:9.2.45] Use Exercise [exer:9.2.44} to show that a function \(y=y(x)\) satisfies the equation \[a_0x^3y'''+a_1x^2y''+a_2xy'+a_3y=0, \tag{A}\] on \((0,\infty)\) if and only if the function \(Y(t)=y(e^t)\) satisfies \[a_0{d^3Y\over dt^3}+(a_1-3a_0) {d^2Y\over dt^2}+(a_2-a_1+2a_0) {dY\over dt}+a_3Y=0.\nonumber\] Assuming that \(a_0\), \(a_1\), \(a_2\), \(a_3\) are real and \(a_0 \ne0\), find the possible forms for the general solution of (A).