Saltar al contenido principal
LibreTexts Español

8.7E: Ecuaciones de Coeficiente Constante con Impulsos (Ejercicios)

  1. Última actualización
  2. Guardar como PDF
  • Page ID
    114853

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Q8.7.1

    En Ejercicios 8.7.1-8.7.20 resolver el problema de valor inicial. Grafique la solución para los Ejercicios 8.7.2, 8.7.4, 8.7.9 y 8.7.19.

    1. \(y''+3y'+2y=6e^{2t}+2\delta(t-1), \quad y(0)=2,\quad y'(0)=-6\)

    2. \(y''+y'-2y=-10e^{-t}+5\delta(t-1), \quad y(0)=7,\quad y'(0)=-9\)

    3. \(y''-4y=2e^{-t}+5\delta(t-1), \quad y(0)=-1,\quad y'(0)=2\)

    4. \(y''+y=\sin3t+2\delta(t-\pi/2), \quad y(0)=1,\quad y'(0)=-1\)

    5. \(y''+4y=4+\delta(t-3\pi), \quad y(0)=0,\quad y'(0)=1\)

    6. \(y''-y=8+2\delta(t-2), \quad y(0)=-1,\quad y'(0)=1\)

    7. \(y''+y'=e^t+3\delta(t-6), \quad y(0)=-1,\quad y'(0)=4\)

    8. \(y''+4y=8e^{2t}+\delta(t-\pi/2), \quad y(0)=8,\quad y'(0)=0\)

    9. \(y''+3y'+2y=1+\delta(t-1), \quad y(0)=1,\quad y'(0)=-1\)

    10. \(y''+2y'+y=e^t+2\delta(t-2), \quad y(0)=-1,\quad y'(0)=2\)

    11. \(y''+4y=\sin t+\delta(t-\pi/2), \quad y(0)=0,\quad y'(0)=2\)

    12. \(y''+2y'+2y=\delta(t-\pi)-3\delta(t-2\pi), \quad y(0)=-1,\quad y'(0)=2\)

    13. \(y''+4y'+13y=\delta(t-\pi/6)+2\delta(t-\pi/3), \quad y(0)=1,\quad y'(0)=2\)

    14. \(2y''-3y'-2y=1+\delta(t-2), \quad y(0)=-1,\quad y'(0)=2\)

    15. \(4y''-4y'+5y=4\sin t-4\cos t+\delta(t-\pi/2)-\delta(t-\pi), \quad y(0)=1,\quad y'(0)=1\)

    16. \(y''+y=\cos2t+2\delta(t-\pi/2)-3\delta(t-\pi), \quad y(0)=0,\quad y'(0)=-1\)

    17. \(y''-y=4e^{-t}-5\delta(t-1)+3\delta(t-2), \quad y(0)=0,\quad y'(0)=0\)

    18. \(y''+2y'+y=e^t-\delta(t-1)+2\delta(t-2), \quad y(0)=0,\quad y'(0)=-1\)

    19. \(y''+y=f(t)+\delta(t-2\pi), \quad y(0)=0,\quad y'(0)=1\),

    \(f(t)=\left\{\begin{array}{cl} \sin2t,&0\le t<\pi,\\[4pt]0,&t\ge \pi.\end{array}\right.\)

    20. \(y''+4y=f(t)+\delta(t-\pi)-3\delta(t-3\pi/2), \quad y(0)=1,\quad y'(0)=-1\),

    \(f(t)=\left\{\begin{array}{cl}1,&0\le t<\pi/2,\\[4pt]2,&t\ge \pi/2\end{array}\right.\)

    Q8.7.2

    21. \(y''+y=\delta(t), \quad y(0)=1,\quad y_-'(0)=-2\)

    22. \(y''-4y=3\delta(t), \quad y(0)=-1,\quad y_-'(0)=7\)

    23. \(y''+3y'+2y=-5\delta(t), \quad y(0)=0,\quad y_-'(0)=0\)

    24. \(y''+4y'+4y=-\delta(t), \quad y(0)=1,\quad y_-'(0)=5\)

    25. \(4y''+4y'+y=3\delta(t), \quad y(0)=1,\quad y_-'(0)=-6\)

    Q8.7.3

    En Ejercicios 8.7.26-8.7.28, resolver el problema de valor inicial\[ay_{h}'' + by_{h}'+cy_{h}=\left\{\begin{array}{ll} {0,}&{0\leq t<t_{0}}\\{1/h, }&{t_{0}\leq t< t_{0} +h, }\\{0,}&{t\geq t_{0}+h, } \end{array} \right. \quad y_{h}(0)=0, y_{h}'(0)=0\nonumber \] donde\(t_{0}>0\) y\(h>0\). Luego encuentra\[w=\cal{L}^{-1}\left(\frac{1}{as^{2}+bs+c} \right) \nonumber \] y verifica el Teorema 8.7.1 graficando\(w\) y\(y_{h}\) en los mismos ejes, para pequeños valores positivos de\(h\).

    26. \(y''+2y'+2y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)

    27. \(y''+2y'+y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)

    28. \(y''+3y'+2y=f_h(t), \quad y(0)=0,\quad y'(0)=0\)

    Q8.7.4

    29. Recordemos de la Sección 6.2 que el desplazamiento de un objeto de masa\(m\) en un sistema de masa de resorte en oscilación amortiguada libre es

    \[my''+cy'+ky=0, \quad y(0)=y_0,\quad y'(0)=v_0,\nonumber \]

    y que\(y\) se puede escribir como

    \[y=Re^{-ct/2m}\cos(\omega_1t-\phi)\nonumber\]

    si el movimiento está subamortiguado. Supongamos\(y(\tau)=0\). Encuentra el impulso que habría que aplicar al objeto\(t=\tau\) para ponerlo en equilibrio.

    30. Resolver el problema de valor inicial. Encuentra una fórmula que no implique funciones de paso y que represente\(y\) en cada subintervalo\([0,\infty)\) en el que la función de forzamiento es cero.

    1. \(y''-y=\sum_{k=1}^\infty\delta(t-k), \quad y(0)=0,\quad y'(0)=1\)
    2. \(y''+y=\sum_{k=1}^\infty\delta(t-2k\pi), \quad y(0)=0,\quad y'(0)=1\)
    3. \(y''-3y'+2y=\sum_{k=1}^\infty\delta(t-k), \quad y(0)=0,\quad y'(0)=1\)
    4. \(y''+y=\sum_{k=1}^\infty\delta(t-k\pi), \quad y(0)=0,\quad y'(0)=0\)

    This page titled 8.7E: Ecuaciones de Coeficiente Constante con Impulsos (Ejercicios) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.