8.5E: Constant Coefficient Equations with Piecewise Continuous Forcing Functions (Exercises)
( \newcommand{\kernel}{\mathrm{null}\,}\)
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={3,0≤t<π,0,t≥π,y(0)=0,y′(0)=0
2. y″+y={3,0≤t<4,;2t−5,t>4,y(0)=1,y′(0)=0
3. y″−2y′={4,0≤t<1,6,t≥1,y(0)=−6,y′(0)=1
4. y″−y={e2t,0≤t<2,1,t≥2,y(0)=3,y′(0)=−1
5. y″−3y′+2y={0,0≤t<1,1,1≤t<2,−1,t≥2,y(0)=−3,y′(0)=1
6. y″+4y={|sint|,0≤t<2π,0,t≥2π,y(0)=−3,y′(0)=1
7. y″−5y′+4y={1,0≤t<1−1,1≤t<2,0,t≥2,y(0)=3,y′(0)=−5
8. y″+9y={cost,0≤t<3π2,sint,t≥3π2,y(0)=0,y′(0)=0
9. y″+4y={t,0≤t<π2,π,t≥π2,y(0)=0,y′(0)=0
10. y″+y={−t,0≤t<π,−t,t≥π,y(0)=0,y′(0)=0
11. y″−3y′+2y={0,0≤t<2,2t−4,t≥2,,y(0)=0,y′(0)=0
12. y″+y={t,0≤t<2π,−2t,t≥2π,y(0)=1,y′(0)=2
13. y″+3y′+2y={−1,0≤t<2,−1,t≥2,y(0)=0,y′(0)=0
14. y″−4y′+3y={−1,0≤t<1,−1,t≥1,y(0)=0,y′(0)=0
15. y″+2y′+y={et,0≤t<1,et−1,t≥1,y(0)=3,y′(0)=−1
16. y″+2y′+y={4et,0≤t<1,0,t≥1,y(0)=0,y′(0)=0
17. y″+3y′+2y={e−t,0≤t<1,0,t≥1,y(0)=1,y′(0)=−1
18. y″−4y′+4y={e2t,0≤t<2,−e2t,t≥2,y(0)=0,y′(0)=−1
19. y″={t2,0≤t<1,−t,1≤t<2,t+1,t≥2,y(0)=1,y′(0)=0
20. y″+2y′+2y={1,0≤t<2π,t,2π≤t<3π,−1,t≥3π,y(0)=2,y′(0)=−1
Q8.5.2
21. Solve the initial value problemy″=f(t),y(0)=0,y′(0)=0,wheref(t)=m+1,m≤t<m+1,m=0,1,2,….
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),y(0)=0,y′(0)=0;
f(t)=m+1,mπ≤t<(m+1)π,m=0,1,2,…. - y″+y=f(t),y(0)=0,y′(0)=0;
f(t)=(m+1)t,2mπ≤t<2(m+1)π,m=0,1,2,… HINT: You'll need the formula 1+2+⋯+m=m(m+1)2. - y″+y=f(t),y(0)=0,y′(0)=0;
f(t)=(−1)m,mπ≤t<(m+1)π,m=0,1,2,…. - y″−y=f(t),y(0)=0,y′(0)=0;
f(t)=m+1,m≤t<(m+1),m=0,1,2,….
HINT: You will need the formula 1+r+...+rm=1−rm+11−r(r≠1). - y″+2y′+2y=f(t),y(0)=0,y′(0)=0;
f(t)=(m+1)(sint+2cost),2mπ≤t<2(m+1)π,m=0,1,2,….
(See the hint in d.) - y″−3y′+2y=f(t),y(0)=0,y′(0)=0;
- f(t)=m+1,m≤t<m+1,m=0,1,2,….
(See the hints in b and d.)
23.
- Let g be continuous on (α,β) and differentiable on the (α,t0) and (t0,β). Suppose A=limt→t0−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
- 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).
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) \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
- 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.