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

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## Q8.5.1

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. $$y''+y=\left\{\begin{array}{cl} 3,&0\le t<4,\\; 2t-5,&t > 4,\end{array}\right.\qquad y(0)=1,\quad y'(0)=0$$ 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. $$y''+9y=\left\{\begin{array}{ll}{\cos t,}&{0\leq t<\frac{3\pi }{2},}\\{\sin t,}&{t\geq \frac{3\pi }{2},} \end{array} \right. \quad y(0)=0,\: y'(0)=0$$ 9. $$y''+4y=\left\{\begin{array}{ll}{t,}&{0\leq t<\frac{\pi }{2},}\\{\pi ,}&{t\geq \frac{\pi }{2},} \end{array} \right. \quad y(0)=0,\: y'(0)=0$$ 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. $$y''-3y'+2y=\left\{\begin{array}{cl} 0,&0\le t<2,\\2t-4,&t\ge 2,\end{array}\right. ,\quad y(0)=0,\quad y'(0)=0$$ 12. $$y''+y=\left\{\begin{array}{cl} t,&0\le t<2\pi,\\-2t,&t\ge 2\pi,\end{array}\right.\quad y(0)=1,\quad y'(0)=2$$ 13. $$y''+3y'+2y=\left\{\begin{array}{cl}\phantom{-}1,&0\le t<2,\\-1,&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=0$$ 14. $$y''-4y'+3y=\left\{\begin{array}{cl}-1,&0\le t<1,\\\phantom{-}1,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0$$ 15. $$y''+2y'+y=\left\{\begin{array}{cl} e^t,&0\le t<1,\\e^t-1,&t\ge 1,\end{array}\right.\; y(0)=3,\; y'(0)=-1$$ 16. $$y''+2y'+y=\left\{\begin{array}{cl} 4e^t,&0\le t<1,\\0,&t\ge 1,\end{array}\right.\; y(0)=0,\; y'(0)=0$$ 17. $$y''+3y'+2y=\left\{\begin{array}{cl} e^{-t},&0\le t<1,\\0,&t\ge 1,\end{array}\right.\; y(0)=1,\; y'(0)=-1$$ 18. $$y''-4y'+4y=\left\{\begin{array}{rl} e^{2t},&0\le t<2,\\-e^{2t},&t\ge 2,\end{array}\right.\; y(0)=0,\; y'(0)=-1$$ 19. $$y''=\left\{\begin{array}{cl}t^2,&0\le t<1,\\-t,&1\le t<2,\\t+1,&t\ge 2,\end{array}\right.\; y(0)=1,\; y'(0)=0$$ 20. $$y''+2y'+2y=\left\{\begin{array}{rl}1,&0\le t<2\pi,\\t,&2\pi\le t<3\pi,\\-1,&t\ge 3\pi,\end{array}\right.\; y(0)=2,\quad y'(0)=-1$$ ## Q8.5.2 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.)

23.

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 $$(\alpha,\beta)$$.

24.

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 $f(t)= \left\{\begin{array}{cl} f_m(t),&t_m\le t< t_{m+1},\quad m=1,\dots,n-1,\\ f_n(t),&t\ge t_n. \end{array}\right.\nonumber$ 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

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

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 $$m=1,\dots,n$$.