13.5: Ecuaciones diferenciales
- Page ID
- 109783
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Método de encubrimiento
Vamos a usar fracciones parciales y el método de encubrimiento. Supondremos que ha visto fracciones parciales. Si no los recuerdas bien o nunca has visto el método de encubrimiento.
Resolver\(y'' - y = e^{2t}\)\(y(0) = 1\),,\(y'(0) = 1\) usando la transformación de Laplace.
Solución
Llamar\(\mathcal{L} (y) = Y\). Aplicar la transformada de Laplace a la ecuación da
\[(s^2 Y - sy(0) - y'(0)) - Y = \dfrac{1}{s - 2}\nonumber \]
Un poco de álgebra ahora da
\[(s^2 - 1) Y = \dfrac{1}{s - 2} + s + 1.\nonumber \]
Entonces
\[Y = \dfrac{1}{(s - 2)(s^2 - 1)} + \dfrac{s + 1}{s^2 - 1} = \dfrac{1}{(s - 2)(s^2 - 1)} + \dfrac{1}{s - 1}\nonumber \]
Usar fracciones parciales para escribir
\[Y = \dfrac{A}{s - 2} + \dfrac{B}{s - 1} + \dfrac{C}{s + 1} + \dfrac{1}{s - 1}.\nonumber \]
El método de encubrimiento da\(A = 1/3, B = -1/2, C = 1/6.\)
Reconocemos
\[\dfrac{1}{s - a}\nonumber \]
como la transformación de Laplace de\(e^{at}\), así
\[y(t) = Ae^{2t} + Be^t + Ce^{-t} + e^t = \dfrac{1}{3} e^{2t} - \dfrac{1}{2} e^t + \dfrac{1}{6} e^{-t} + e^t.\nonumber \]
Resolver\(y'' - y = 1\)\(y(0) = 0\),,\(y'(0) = 0\).
Solución
El resto (cero) las condiciones iniciales son agradables porque no agregarán ningún término al álgebra. Al igual que en el ejemplo anterior aplicamos la transformada de Laplace a toda la ecuación.
\[s^2 Y - Y = \dfrac{1}{s}, \text{ so } Y = \dfrac{1}{s(s^2 - 1)} = \dfrac{1}{s(s - 1)(s+1)} = \dfrac{A}{s} + \dfrac{B}{s -1} + \dfrac{C}{s + 1}\nonumber \]
El método de encubrimiento da\(A = -1, B = 1/2, C = 1/2\). Entonces,
\[y = A + Be^t + Ce^{-t} = -1 + \dfrac{1}{2} e^t + \dfrac{1}{2} e^{-t}.\nonumber \]