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Mathematics LibreTexts

8.5E: Constant Coefficient Equations with Piecewise Continuous Forcing Functions (Exercises)

  • Page ID
    18293
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    In Exercises [exer:8.5.1}– [exer:8.5.20} use the Laplace transform to solve the initial value problem. Where indicated by
    , graph the solution.

    [exer:8.5.1] \(\{y''+y=\left\{\begin{array}{cl} 3,& 0\le t<\pi,\\[5pt] 0,&t\ge\pi,\end{array}\right. \qquad y(0)=0, \quad y'(0)=0}\)

    [exer:8.5.2] \(\{y''+y=\left\{\begin{array}{cl} 3,&0\le t<4,\\; 2t-5,&t > 4,\end{array}\right.\qquad y(0)=1,\quad y'(0)=0}\)

    [exer:8.5.3] \(\{y''-2y'= \left\{\begin{array}{cl} 4,&0\le t<1,\\[5pt] 6,&t\ge 1,\end{array}\right.\qquad y(0)=-6,\quad y'(0)=1 }\)

    [exer:8.5.4] \(\{y''-y=\left\{\begin{array}{cl} e^{2t},&0\le t< 2,\\[5pt] 1,&t\ge 2,\end{array}\right.\qquad y(0)=3,\quad y'(0)=-1 }\)

    [exer:8.5.5] \(\{y''-3y'+2y= \left\{\begin{array}{rl} 0,&0\le t<1,\\[5pt] 1,&1\le t<2,\\[5pt]-1,&t\ge 2, \end{array}\right.\qquad y(0)=-3,\quad y'(0)=1}\)

    [exer:8.5.6] \(\{y''+4y= \left\{\begin{array}{cl}|\sin t|,&0\le t<2\pi,\\[5pt] 0,&t\ge 2\pi,\end{array}\right.\qquad y(0)=-3,\quad y'(0)=1}\)

    [exer:8.5.7] \(\{y''-5y'+4y= \left\{\begin{array}{rl} 1,&0\le t<1\\[5pt] -1,&1\le t<2,\\[5pt] 0,&t\ge 2,\end{array}\right.\qquad y(0)=3,\quad y'(0)=-5}\)

    [exer:8.5.8] \(\{y''+9y= \left\{\begin{array}{cl} \cos t,&0\le t<\

    ParseError: EOF expected (click for details)
    Callstack:
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    ,\\[5pt] \sin t,&t\ge\
    ParseError: EOF expected (click for details)
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    ,\end{array}\right.\qquad y(0)=0,\; y'(0)=0}\)

    [exer:8.5.9] \(\{y''+4y= \left\{\begin{array}{cl} t,&0\le t<\{\pi \over2},\\[5pt]\pi,&t\ge\{ {\pi\over2}}, \end{array}\right.\quad y(0)=0,\quad y'(0)=0}\)

    [exer:8.5.10] \(\{y''+y=\left\{\begin{array}{cl}\phantom{-}t,&0\le t<\pi, \\[5pt]-t,&t\ge\pi ,\end{array}\right.\; y(0)=0,\; y'(0)=0}\)

    [exer:8.5.11] \(\{y''-3y'+2y=\left\{\begin{array}{cl} 0,&0\le t<2,\\2t-4,&t\ge 2,\end{array}\right. ,\quad y(0)=0,\quad y'(0)=0}\)

    [exer:8.5.12] \(\{y''+y=\left\{\begin{array}{cl} t,&0\le t<2\pi,\\-2t,&t\ge 2\pi,\end{array}\right.\quad y(0)=1,\quad y'(0)=2}\)

    [exer:8.5.13] \(\{y''+3y'+2y=\left\{\begin{array}{cl}\phantom{-}1,&0\le t<2,\\-1,&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=0}\)

    [exer:8.5.14] \(\{y''-4y'+3y=\left\{\begin{array}{cl}-1,&0\le t<1,\\\phantom{-}1,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0}\)

    [exer:8.5.15] \(\{y''+2y'+y=\left\{\begin{array}{cl} e^t,&0\le t<1,\\e^t-1,&t\ge 1,\end{array}\right.\; y(0)=3,\; y'(0)=-1}\)

    [exer:8.5.16] \(\{y''+2y'+y=\left\{\begin{array}{cl} 4e^t,&0\le t<1,\\0,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0}\)

    [exer:8.5.17] \(\{y''+3y'+2y=\left\{\begin{array}{cl} e^{-t},&0\le t<1,\\0,&t\ge 1,\end{array}\right.\; y(0)=1,\; y'(0)=-1}\)

    [exer:8.5.18] \(\{y''-4y'+4y=\left\{\begin{array}{rl} e^{2t},&0\le t<2,\\-e^{2t},&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=-1}\)

    [exer:8.5.19] \(\{y''=\left\{\begin{array}{cl}t^2,&0\le t<1,\\-t,&1\le t<2,\\t+1,&t\ge 2,\end{array}\right.\; y(0)=1,\; y'(0)=0}\)

    [exer:8.5.20] \(\{y''+2y'+2y=\left\{\begin{array}{rl}1,&0\le t<2\pi,\\t,&2\pi\le t<3\pi,\\-1,&t\ge 3\pi,\end{array}\right.\; y(0)=2,\quad y'(0)=-1}\)

    [exer:8.5.21] Solve the initial value problem\[y''=f(t), \quad y(0)=0,\quad y'(0)=0,\]where\[f(t)=m+1,\quad m\le t<m+1,\quad m=0,1,2,\dots.\]

    [exer:8.5.22] 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 \(f\).

    \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);

    \(f(t)=m+1,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots\).

    \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);

    \(f(t)=(m+1)t, \quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots\)

    \[1+2+\cdots+m={m(m+1)\over2}.\]

    \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);

    \(f(t)=(-1)^m,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots.\)

    \(y''-y=f(t), \quad y(0)=0,\quad y'(0)=0\);

    \(f(t)=m+1,\quad m\le t<(m+1),\quad m=0,1,2,\dots.\)

    \(y''+2y'+2y=f(t), \quad y(0)=0,\quad y'(0)=0\);

    \(f(t)=(m+1)(\sin t+2\cos t),\quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots.\)

    (See the hint in d.)

    \(y''-3y'+2y=f(t), \quad y(0)=0,\quad y'(0)=0\);

    \(f(t)=m+1,\quad m\le t<m+1,\quad m=0,1,2,\dots.\)

    (See the hints in b and d.)

    [exer:8.5.23] 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 \[\lim_{t\to t_0-}{g(t)-g(t_0)\over t-t_0}=A\quad\mbox{ and }\quad \lim_{t\to t_0+}{g(t)-g(t_0)\over t-t_0}=B.\]

    Conclude from

    a

    that \(g'(t_0)\) exists and \(g'\) is continuous at \(t_0\) if \(A=B\).

    Conclude from

    a

    that if \(g\) is differentiable on \((\alpha,\beta)\) then \(g'\) can’t have a jump discontinuity on \((\alpha,\beta)\).

    [exer:8.5.24]

    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

    \[ay''+by'+cy=f(t) \eqno{\rm (A)}\]

    on \((\alpha,t_0)\) and \((t_0,\beta)\). Show that

    \[y''(t_0+)-y''(t_0-)={f(t_0+)-f(t_0-)\over a}\ne0.\]

    Use

    a

    and Exercise [exer:8.5.23}

    c

    to show that (A) does not have solutions on any interval \((\alpha,\beta)\) that contains a jump discontinuity of \(f\).

    [exer:8.5.25] 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 \[P_0(t)y''+P_1(t)y'+P_2(t)y=F(t)\]has no solutions on \((a,b)\).

    [exer:8.5.26] Let \(0=t_0<t_1<\cdots <t_n\). Suppose \(f_m\) is continuous on \([t_m,\infty)\) for \(m=1,\dots,n\). Let \[f(t)= \left\{\begin{array}{cl} f_m(t),&t_m\le t< t_{m+1},\quad m=1,\dots,n-1,\\ f_n(t),&t\ge t_n. \end{array}\right.\] Show that the solution of

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

    as defined following Theorem 8.5.1, is given by

    \[y=\left\{\begin{array}{cl} z_0(t),&0\le t<t_1,\\[5pt] z_0(t)+ z_1(t),&t_1\le t<t_2,\\ &\vdots\\ z_0+\cdots+z_{n-1}(t),&t_{n-1}\le t<t_n,\\[5pt] z_0+\cdots+ z_n(t),&t\ge t_n, \end{array}\right.\]

    where \(z_0\) is the solution of

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

    and \(z_m\) is the solution of

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

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