7.5E: Constant Coefficient Equations with Piecewise Continuous Forcing Functions (Exercises)
- Page ID
- 43331
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Q7.5.1
In Exercises 7.5.1-7.5.20 use the Laplace transform to solve the initial value problem. Graph the solution for Exercise 7.5.6, 7.5.9, 7.5.13, and 7.5.19.
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. \(y''+2y'+2y=\left\{\begin{array}{rl}1,&0\le t<2\pi,\\t,&2\pi\le t<3\pi,\\-1,&t\ge 3\pi,\end{array}\right.\; y(0)=2,\quad y'(0)=-1\)
Q7.5.2
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 \(f\).
- \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);
\(f(t)=m+1,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots\). - \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);
\(f(t)=(m+1)t, \quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots\) HINT: You'll need the formula \[1+2+\cdots+m={m(m+1)\over2}.\nonumber \] - \(y''+y=f(t), \quad y(0)=0,\quad y'(0)=0\);
\(f(t)=(-1)^m,\quad m\pi\le t<(m+1)\pi,\quad m=0,1,2,\dots.\) - \(y''-y=f(t), \quad y(0)=0,\quad y'(0)=0\);
\(f(t)=m+1,\quad m\le t<(m+1),\quad m=0,1,2,\dots.\)
HINT: You will need the formula \[1+r+...+r^{m}=\frac{1-r^{m+1}}{1-r}(r\neq 1).\nonumber \] - \(y''+2y'+2y=f(t), \quad y(0)=0,\quad y'(0)=0\);
\(f(t)=(m+1)(\sin t+2\cos t),\quad 2m\pi\le t<2(m+1)\pi,\quad m=0,1,2,\dots.\)
(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.)