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8.5.1: Constant Coefficient Equations with Piecewise Continuous Forcing Functions (Exercises)

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    In Exercises 8.5.1-8.5.20 use the Laplace transform to solve the initial value problem. Graph the solution for Exercise 8.5.6, 8.5.9, 8.5.13, and 8.5.19.

    1. \(y''+y=\left\{\begin{array}{cl} 3,& 0\le t<\pi,\\[4pt] 0,&t\ge\pi,\end{array}\right. \qquad y(0)=0, \quad y'(0)=0\)

    2. \(y''+y=\left\{\begin{array}{cl} 3,&0\le t<4,\\[4pt]; 2t-5,&t > 4,\end{array}\right.\qquad y(0)=1,\quad y'(0)=0\)

    3. \(y''-2y'= \left\{\begin{array}{cl} 4,&0\le t<1,\\[4pt] 6,&t\ge 1,\end{array}\right.\qquad y(0)=-6,\quad y'(0)=1 \)

    4. \(y''-y=\left\{\begin{array}{cl} e^{2t},&0\le t< 2,\\[4pt] 1,&t\ge 2,\end{array}\right.\qquad y(0)=3,\quad y'(0)=-1 \)

    5. \(y''-3y'+2y= \left\{\begin{array}{rl} 0,&0\le t<1,\\[4pt] 1,&1\le t<2,\\[4pt]-1,&t\ge 2, \end{array}\right.\qquad y(0)=-3,\quad y'(0)=1\)

    6. \(y''+4y= \left\{\begin{array}{cl}|\sin t|,&0\le t<2\pi,\\[4pt] 0,&t\ge 2\pi,\end{array}\right.\qquad y(0)=-3,\quad y'(0)=1\)

    7. \(y''-5y'+4y= \left\{\begin{array}{rl} 1,&0\le t<1\\[4pt] -1,&1\le t<2,\\[4pt] 0,&t\ge 2,\end{array}\right.\qquad y(0)=3,\quad y'(0)=-5\)

    8. \(y''+9y=\left\{\begin{array}{ll}{\cos t,}&{0\leq t<\frac{3\pi }{2},}\\[4pt]{\sin t,}&{t\geq \frac{3\pi }{2},} \end{array} \right. \quad y(0)=0,\: y'(0)=0 \)

    9. \(y''+4y=\left\{\begin{array}{ll}{t,}&{0\leq t<\frac{\pi }{2},}\\[4pt]{\pi ,}&{t\geq \frac{\pi }{2},} \end{array} \right. \quad y(0)=0,\: y'(0)=0 \)

    10. \(y''+y=\left\{\begin{array}{cl}\phantom{-}t,&0\le t<\pi, \\[4pt]-t,&t\ge\pi ,\end{array}\right.\; y(0)=0,\; y'(0)=0\)

    11. \(y''-3y'+2y=\left\{\begin{array}{cl} 0,&0\le t<2,\\[4pt]2t-4,&t\ge 2,\end{array}\right. ,\quad y(0)=0,\quad y'(0)=0\)

    12. \(y''+y=\left\{\begin{array}{cl} t,&0\le t<2\pi,\\[4pt]-2t,&t\ge 2\pi,\end{array}\right.\quad y(0)=1,\quad y'(0)=2\)

    13. \(y''+3y'+2y=\left\{\begin{array}{cl}\phantom{-}1,&0\le t<2,\\[4pt]-1,&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=0\)

    14. \(y''-4y'+3y=\left\{\begin{array}{cl}-1,&0\le t<1,\\[4pt]\phantom{-}1,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0\)

    15. \(y''+2y'+y=\left\{\begin{array}{cl} e^t,&0\le t<1,\\[4pt]e^t-1,&t\ge 1,\end{array}\right.\; y(0)=3,\; y'(0)=-1\)

    16. \(y''+2y'+y=\left\{\begin{array}{cl} 4e^t,&0\le t<1,\\[4pt]0,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0\)

    17. \(y''+3y'+2y=\left\{\begin{array}{cl} e^{-t},&0\le t<1,\\[4pt]0,&t\ge 1,\end{array}\right.\; y(0)=1,\; y'(0)=-1\)

    18. \(y''-4y'+4y=\left\{\begin{array}{rl} e^{2t},&0\le t<2,\\[4pt]-e^{2t},&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=-1\)

    19. \(y''=\left\{\begin{array}{cl}t^2,&0\le t<1,\\[4pt]-t,&1\le t<2,\\[4pt]t+1,&t\ge 2,\end{array}\right.\; y(0)=1,\; y'(0)=0\)

    20. \(y''+2y'+2y=\left\{\begin{array}{rl}1,&0\le t<2\pi,\\[4pt]t,&2\pi\le t<3\pi,\\[4pt]-1,&t\ge 3\pi,\end{array}\right.\; y(0)=2,\quad y'(0)=-1\)


    21. Solve the initial value problem\[y''=f(t), \quad y(0)=0,\quad y'(0)=0,\nonumber \]where\[f(t)=m+1,\quad m\le t<m+1,\quad m=0,1,2,\dots.\nonumber \]

    22. Solve the given initial value problem and find a formula that does not involve step functions and represents \(y\) on each interval of continuity of \(f\).

    1. \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);
      \(f(t)=m+1,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots\).
    2. \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);
      \(f(t)=(m+1)t, \quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots\) HINT: You'll need the formula \[1+2+\cdots+m={m(m+1)\over2}.\nonumber \]
    3. \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);
      \(f(t)=(-1)^m,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots.\)
    4. \(y''-y=f(t), \quad y(0)=0,\quad y'(0)=0\);
      \(f(t)=m+1,\quad m\le t<(m+1),\quad m=0,1,2,\dots.\)
      HINT: You will need the formula \[1+r+...+r^{m}=\frac{1-r^{m+1}}{1-r}(r\neq 1).\nonumber \]
    5. \(y''+2y'+2y=f(t), \quad y(0)=0,\quad y'(0)=0\);
      \(f(t)=(m+1)(\sin t+2\cos t),\quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots.\)
      (See the hint in d.)
    6. \(y''-3y'+2y=f(t), \quad y(0)=0,\quad y'(0)=0\);
    7. \(f(t)=m+1,\quad m\le t<m+1,\quad m=0,1,2,\dots.\)
      (See the hints in b and d.)


    1. Let \(g\) be continuous on \((\alpha,\beta)\) and differentiable on the \((\alpha,t_0)\) and \((t_0,\beta)\). Suppose \(A=\lim_{t\to t_0-}g'(t)\) and \(B=\lim_{t\to t_0+}g'(t)\) both exist. Use the mean value theorem to show that \[\lim_{t\to t_0-}{g(t)-g(t_0)\over t-t_0}=A\quad\mbox{ and }\quad \lim_{t\to t_0+}{g(t)-g(t_0)\over t-t_0}=B.\nonumber \]
    2. Conclude from (a) that \(g'(t_0)\) exists and \(g'\) is continuous at \(t_0\) if \(A=B\).
    3. Conclude from (a) that if \(g\) is differentiable on \((\alpha,\beta)\) then \(g'\) can’t have a jump discontinuity on \((\alpha,\beta)\).


    1. Let \(a\), \(b\), and \(c\) be constants, with \(a\ne0\). Let \(f\) be piecewise continuous on an interval \((\alpha,\beta)\), with a single jump discontinuity at a point \(t_0\) in \((\alpha,\beta)\). Suppose \(y\) and \(y'\) are continuous on \((\alpha,\beta)\) and \(y''\) on \((\alpha,t_0)\) and \((t_0,\beta)\). Suppose also that \[ay''+by'+cy=f(t) \tag{A} \] on \((\alpha,t_0)\) and \((t_0,\beta)\). Show that \[y''(t_0+)-y''(t_0-)={f(t_0+)-f(t_0-)\over a}\ne0.\nonumber \]
    2. Use (a) and Exercise 8.5.23c to show that (A) does not have solutions on any interval \((\alpha,\beta)\) that contains a jump discontinuity of \(f\).

    25. Suppose \(P_0,P_1\), and \(P_2\) are continuous and \(P_0\) has no zeros on an open interval \((a,b)\), and that \(F\) has a jump discontinuity at a point \(t_0\) in \((a,b)\). Show that the differential equation \[P_0(t)y''+P_1(t)y'+P_2(t)y=F(t)\nonumber \]has no solutions on \((a,b)\). HINT: Generalize the result of Exercise 8.5.24 and use Exercise 8.5.23c.

    26. Let \(0=t_0<t_1<\cdots <t_n\). Suppose \(f_m\) is continuous on \([t_m,\infty)\) for \(m=1,\dots,n\). Let \[f(t)= \left\{\begin{array}{cl} f_m(t),&t_m\le t< t_{m+1},\quad m=1,\dots,n-1,\\[4pt] f_n(t),&t\ge t_n. \end{array}\right.\nonumber \] Show that the solution of

    \[ay''+by'+cy=f(t), \quad y(0)=k_0,\quad y'(0)=k_1,\nonumber \]

    as defined following Theorem 8.5.1, is given by

    \[y=\left\{\begin{array}{cl} z_0(t),&0\le t<t_1,\\[4pt] z_0(t)+ z_1(t),&t_1\le t<t_2,\\[4pt] &\vdots\\[4pt] z_0+\cdots+z_{n-1}(t),&t_{n-1}\le t<t_n,\\[4pt] z_0+\cdots+ z_n(t),&t\ge t_n, \end{array}\right.\nonumber \]

    where \(z_0\) is the solution of

    \[az''+bz'+cz=f_0(t), \quad z(0)=k_0,\quad z'(0)=k_1\nonumber \]

    and \(z_m\) is the solution of

    \[az''+bz'+cz=f_m(t)-f_{m-1}(t), \quad z(t_m)=0,\quad z'(t_m)=0\nonumber \]

    for \(m=1,\dots,n\).

    This page titled 8.5.1: Constant Coefficient Equations with Piecewise Continuous Forcing Functions (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.