Solución de ecuaciones diferenciales ordinarias con coeficientes constantes utilizando la transformada de Laplace. Laplace.

Introducción

La transformada de Laplace presenta gran utilidad para resolver ecuaciones diferenciales con coeficientes constantes. Considerando como ejemplo, una ecuación diferencial de segundo orden

$\displaystyle \frac{d^2y}{dt^2} + \alpha \frac{dy}{dt} + \beta y = f(t)$

donde $\alpha$ y $\beta$ son constantes sometidas a ciertas condiciones iniciales o condiciones de la frontera como lo son $y(0) = A$ y $y'(0) = B$ donde A y B son constantes. Aplicando la transformada de Laplace en ambos miembros se obtiene una ecuación algebraica para determinar $\mathcal{L}[y(t)] = Y(s)$ y finalmente se calcula la transformada inversa de Laplace. Este procedimiento se puede aplicar si la ecuación diferencial es de orden superior.

Problemas resueltos

Problema 1. Resolver $y'' +y = t$ para $y(0) = 1$ y $y'(0) = -2$ .

Solución. De la ecuación diferencial, se aplica la transformada de Laplace por ambos lados.

$\displaystyle \mathcal{L} [y'' +y] = \mathcal{L} [t]$

$\displaystyle \mathcal{L} [y''] + \mathcal{L} [y] = \mathcal{L} [t]$

Aplicando las fórmulas y propiedades correspondientes para cada término

$\displaystyle s^2 Y(s) - s y(0) - y'(0) + Y(s) = \frac{1}{s^2}$

Sabiendo que $y(0)=1$ y $y'(0) = -2$

$\displaystyle s^2 Y(s) - s (1) - (-2) + Y(s) = \frac{1}{s^2}$

$\displaystyle s^2 Y(s) - s + 2 + Y(s) = \frac{1}{s^2}$

Despejando $Y(s)$

$\displaystyle (s^2 +1) Y(s) - s + 2 = \frac{1}{s^2}$

$\displaystyle (s^2 +1) Y(s) = s - 2 + \frac{1}{s^2}$

$\displaystyle Y(s) = \frac{s - 2}{s^2+1} + \frac{1}{s^2 (s^2+1)}$

$\displaystyle Y(s) = \frac{s}{s^2+1} - \frac{2}{s^2+1} + \frac{A}{s^2} + \frac{B}{s^2+1}$

$\displaystyle Y(s) = \frac{s}{s^2+1} - \frac{2}{s^2+1} + \frac{1}{s^2} - \frac{1}{s^2+1}$

Tomando la transformada inversa de Laplace en ambos miembros

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[\frac{s}{s^2+1} - \frac{2}{s^2+1} + \frac{1}{s^2} - \frac{1}{s^2+1} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[\frac{s}{s^2+1} \right] - \mathcal{L}^{-1} \left[\frac{2}{s^2+1} \right] + \mathcal{L}^{-1} \left[\frac{1}{s^2}\right] - \mathcal{L}^{-1} \left[\frac{1}{s^2+1} \right]$

$\displaystyle y(t) = \cos{t} - 2 \sin{t} + \frac{t}{1!} - \sin{t}$

$\displaystyle y(t) = \cos{t} - 3 \sin{t} + t$

Finalmente,

$\displaystyle \therefore y(t) = \cos{t} - 3 \sin{t} + t$

Problema 2. Resolver $y'' - 3y' + 2y = 4e^{2t}$ para $y(0)=-3$ y $y'(0)=5$ .

Solución. De la ecuación diferencial, se toma la transformada de Laplace en ambos lados

$\displaystyle y'' - 3y' + 2y = 4e^{2t}$

$\displaystyle \mathcal{L} [y'' - 3y' + 2y] = \mathcal{L} [4e^{2t}]$

$\displaystyle \mathcal{L} [y''] - \mathcal{L} [3y'] + \mathcal{L} [2y] = \mathcal{L} [4e^{2t}]$

$\displaystyle \mathcal{L} [y''] - 3 \mathcal{L} [y'] + 2\mathcal{L} [y] = 4\mathcal{L} [e^{2t}]$

Aplicando las fórmulas y propiedades para cada término

$\displaystyle \mathcal{L} [y''] - 3 \mathcal{L} [y'] + 2\mathcal{L} [y] = 4\mathcal{L} [e^{2t}]$

$\displaystyle s^2 Y(s) - s y(0) - y'(0) - 3 \left[s Y(s) - y(0) \right] + 2 Y(s) = 4 \left(\frac{1}{s-2} \right)$

Sabiendo que $y(0) = -3$ y $y'(0) = 5$ , resulta

$\displaystyle s^2 Y(s) - s (-3) - (5) - 3 \left[s Y(s) - (-3) \right] + 2 Y(s) = \frac{4}{s-2}$

$\displaystyle s^2 Y(s) + 3s - 5 - 3 \left[s Y(s) + 3 \right] + 2 Y(s) = \frac{4}{s-2}$

$\displaystyle s^2 Y(s) + 3s - 5 - 3s Y(s) - 9 + 2 Y(s) = \frac{4}{s-2}$

$\displaystyle (s^2 - 3s + 2) Y(s) + 3s - 14 = \frac{4}{s-2}$

$\displaystyle (s^2 - 3s + 2) Y(s) =14 - 3s + \frac{4}{s-2}$

$\displaystyle Y(s) =\frac{14 - 3s}{s^2-3s+2} + \frac{4}{(s-2)(s^2-3s+2)}$

$\displaystyle Y(s) =\frac{(s-2)(14 - 3s)}{(s-2)(s^2-3s+2)} + \frac{4}{(s-2)(s^2-3s+2)}$

$\displaystyle Y(s) =\frac{14s-3s^2-28+6s}{(s-2)(s^2-3s+2)} + \frac{4}{(s-2)(s^2-3s+2)}$

$\displaystyle Y(s) =\frac{-3s^2+20s-28}{(s-2)(s^2-3s+2)} + \frac{4}{(s-2)(s^2-3s+2)}$

$\displaystyle Y(s) =\frac{-3s^2+20s-28+4}{(s-2)(s^2-3s+2)}$

$\displaystyle Y(s) =\frac{-3s^2+20s-24}{(s-2)(s^2-3s+2)}$

$\displaystyle Y(s) =\frac{-3s^2+20s-24}{(s-2)(s-1)(s-2)}$

$\displaystyle Y(s) =\frac{-3s^2+20s-24}{(s-2)^2(s-1)}$

Aplicando el método de fracciones parciales,

$\displaystyle Y(s) = \frac{4}{(s-2)^2} + \frac{4}{(s-2)} - \frac{7}{(s-1)}$

Utilizando la transformada inversa de Laplace, resulta

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[\frac{4}{(s-2)^2} \right] + \mathcal{L}^{-1} \left[\frac{4}{(s-2)} \right] - \mathcal{L}^{-1} \left[\frac{7}{(s-1)} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = 4 \mathcal{L}^{-1} \left[\frac{1}{(s-2)^2} \right] + 4 \mathcal{L}^{-1} \left[\frac{1}{(s-2)} \right] - 7 \mathcal{L}^{-1} \left[\frac{1}{(s-1)} \right]$

$\displaystyle y(t) = 4 t e^{2t} + 4 e^{2t} - 7e^{t}$

Así que, el resultado final es

$\displaystyle \therefore y(t) = 4 t e^{2t} + 4 e^{2t} - 7e^{t}$

Problema 3. Resolver $\displaystyle y'' + 2y' + 5y = e^{-t} \sin{t}$ , donde $y(0)=0$ y $y'(0) = 1$ .

Solución. Tomando la transformada de Laplace por ambos lados y aplicando la propiedad de linealidad, resulta que

$\displaystyle y'' + 2y' + 5y = e^{-t} \sin{t}$

$\displaystyle \mathcal{L} [y'' + 2y' + 5y] = \mathcal{L} [e^{-t} \sin{t}]$

$\displaystyle \mathcal{L} [y''] + \mathcal{L} [2y'] + \mathcal{L} [5y] = \mathcal{L} [e^{-t} \sin{t}]$

$\displaystyle \mathcal{L} [y''] + 2\mathcal{L} [y'] + 5\mathcal{L} [y] = \mathcal{L} [e^{-t} \sin{t}]$

$\displaystyle s^2 Y(s) - s y(0) - y'(0) + 2[s Y(s) - y(0)] + 5 Y(s) = \frac{1}{(s+1)^2+1}$

$\displaystyle s^2 Y(s) - s y(0) - y'(0) + 2[s Y(s) - y(0)] + 5 Y(s) = \frac{1}{s^2+2s+1+1}$

$\displaystyle s^2 Y(s) - s y(0) - y'(0) + 2[s Y(s) - y(0)] + 5 Y(s) = \frac{1}{s^2+2s+2}$

Tomando las condiciones iniciales $y(0)=0$ y $y'(0)=1$ , se tiene que

$\displaystyle s^2 Y(s) - s (0) - (1) + 2[s Y(s) - 0] + 5 Y(s) = \frac{1}{s^2+2s+2}$

$\displaystyle s^2 Y(s) - 1 + 2s Y(s) + 5 Y(s) = \frac{1}{s^2+2s+2}$

$\displaystyle (s^2 + 2s + 5) Y(s) = 1+ \frac{1}{s^2+2s+2}$

$\displaystyle (s^2 + 2s + 5) Y(s) = \frac{s^2+2s+3}{s^2+2s+2}$

$\displaystyle Y(s) = \frac{s^2+2s+3}{(s^2+2s+5)(s^2+2s+2)}$

Por el método de fracciones parciales

$\displaystyle Y(s) = \frac{As+B}{s^2+2s+2} + \frac{Cs+D}{s^2+2s+5}$

$\displaystyle Y(s) = \frac{0s+1/3}{s^2+2s+2} + \frac{0s+2/3}{s^2+2s+5}$

$\displaystyle Y(s) = \frac{1/3}{s^2+2s+2} + \frac{2/3}{s^2+2s+5}$

Aplicando $\mathcal{L}^{-1}$ en ambos miembros

$\displaystyle \mathcal{L}^{-1} \left[Y(s) \right] = \mathcal{L}^{-1} \left[ \frac{1/3}{s^2+2s+2} + \frac{2/3}{s^2+2s+5} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[ \frac{1/3}{s^2+2s+2} \right] + \mathcal{L}^{-1} \left[\frac{2/3}{s^2+2s+5} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \frac{1}{3} \mathcal{L}^{-1} \left[ \frac{1}{s^2+2s+2} \right] + \frac{2}{3} \mathcal{L}^{-1} \left[\frac{1}{s^2+2s+5} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \frac{1}{3} \mathcal{L}^{-1} \left[ \frac{1}{s^2+2s+(\frac{2}{2})^2 - (\frac{2}{2})^2 +2} \right] + \frac{2}{3} \mathcal{L}^{-1} \left[\frac{1}{s^2+2s+(\frac{2}{2})^2 - (\frac{2}{2})^2 +5} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \frac{1}{3} \mathcal{L}^{-1} \left[ \frac{1}{s^2+2s+(1)^2 - (1)^2 +2} \right] + \frac{2}{3} \mathcal{L}^{-1} \left[\frac{1}{s^2+2s+(1)^2 - (1)^2 +5} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \frac{1}{3} \mathcal{L}^{-1} \left[ \frac{1}{(s+1)^2 - (1)^2 +2} \right] + \frac{2}{3} \mathcal{L}^{-1} \left[\frac{1}{(s+1)^2 - (1)^2 +5} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \frac{1}{3} \mathcal{L}^{-1} \left[ \frac{1}{(s+1)^2 - 1 +2} \right] + \frac{2}{3} \mathcal{L}^{-1} \left[\frac{1}{(s+1)^2 - 1 +5} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \frac{1}{3} \mathcal{L}^{-1} \left[ \frac{1}{(s+1)^2 +1} \right] + \frac{2}{3} \mathcal{L}^{-1} \left[\frac{1}{(s+1)^2 + 4} \right]$

$\displaystyle y(t) = \frac{1}{3} e^{-t} \sin{t} + \frac{2}{3} \cdot \frac{1}{2} e^{-t} \sin{2t}$

$\displaystyle y(t) = \frac{1}{3} e^{-t} \sin{t} + \frac{1}{3} e^{-t} \sin{2t}$

$\displaystyle y(t) = \frac{1}{3} e^{-t} (\sin{t} + \sin{2t})$

Finalmente

$\displaystyle \therefore y(t) = \frac{1}{3} e^{-t} (\sin{t} + \sin{2t})$

Problema 4. Resolver $\displaystyle y'''-3y''+3y'-y=t^2 e^t$ , donde $y(0)=1$ , $y'(0)=0$ y $y''(0)=-2$ .

Solución. Tomando la transformada de laplace en ambos miembros y aplicando la propiedad de linealidad, resulta que

$\displaystyle y'''-3y''+3y'-y=t^2 e^t$

$\displaystyle \mathcal{L} [y'''-3y''+3y'-y] = \mathcal{L} [t^2 e^t]$

$\displaystyle \mathcal{L} [y'''] - \mathcal{L} [3y''] + \mathcal{L} [3y'] - \mathcal{L} [y] = \mathcal{L} [t^2 e^t]$

$\displaystyle \mathcal{L} [y'''] - 3 \mathcal{L} [y''] + 3 \mathcal{L} [y'] - \mathcal{L} [y] = \mathcal{L} [t^2 e^t]$

$\displaystyle s^3 Y(s) - s^2 y(0) - s y'(0) - y''(0) - 3 [s^2 Y(s) - s y(0) - y'(0)] + 3 [sY(s) - y(0)] - Y(s) = \frac{2!}{(s-1)^3}$

$\displaystyle s^3 Y(s) - s^2 y(0) - s y'(0) - y''(0) - 3 [s^2 Y(s) - s y(0) - y'(0)] + 3 [sY(s) - y(0)] - Y(s) = \frac{2}{(s-1)^3}$

Sabiendo que $y(0)=1$ , $y'(0)=0$ y $y''(0)=-2$ , se tiene lo siguiente

$\displaystyle s^3 Y(s) - s^2 (1) - s (0) - (-2) - 3 [s^2 Y(s) - s (1) - (0)] + 3 [sY(s) - 1] - Y(s) = \frac{2}{(s-1)^3}$

$\displaystyle s^3 Y(s) - s^2 + 2 - 3 [s^2 Y(s) - s] + 3 [sY(s) - 1] - Y(s) = \frac{2}{(s-1)^3}$

$\displaystyle s^3 Y(s) - s^2 + 2 - 3 s^2 Y(s) + 3s + 3s Y(s) - 3 - Y(s) = \frac{2}{(s-1)^3}$

$\displaystyle (s^3-3s^2+3s-1) Y(s) - s^2 + 3s - 1 = \frac{2}{(s-1)^3}$

$\displaystyle (s^3-3s^2+3s-1) Y(s) = s^2 - 3s + 1 + \frac{2}{(s-1)^3}$

$\displaystyle (s-1)^3 Y(s) = s^2 - 2s - s + 1 + 1 - 1 + \frac{2}{(s-1)^3}$

$\displaystyle (s-1)^3 Y(s) = (s^2 - 2s + 1) - (s - 1) - 1 + \frac{2}{(s-1)^3}$

$\displaystyle (s-1)^3 Y(s) = (s+1)^2 + (s - 1) - 1 + \frac{2}{(s-1)^3}$

$\displaystyle Y(s) = \frac{(s+1)^2}{(s-1)^3} + \frac{(s - 1)}{(s-1)^3} - \frac{1}{(s-1)^3} + \frac{2}{(s-1)^3 (s-1)^3}$

$\displaystyle Y(s) = \frac{1}{(s-1)} + \frac{1}{(s-1)^2} - \frac{1}{(s-1)^3} + \frac{2}{(s-1)^6}$

Aplicando $\mathcal{L}^{-1}$ en ambos miembros

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[\frac{1}{(s-1)} + \frac{1}{(s-1)^2} - \frac{1}{(s-1)^3} + \frac{2}{(s-1)^6} \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[\frac{1}{(s-1)} \right] + \mathcal{L}^{-1} \left[\frac{1}{(s-1)^2} \right] - \mathcal{L}^{-1} \left[\frac{1}{(s-1)^3} \right] + \mathcal{L}^{-1} \left[\frac{2}{(s-1)^6} \right]$

$\displaystyle y(t) = e^{t} + \frac{t e^{t}}{1!} - \frac{t^2 e^t}{2!}+ 2 \left(\frac{t^5 e^t}{5!} \right)$

$\displaystyle y(t) = e^{t} + \frac{t e^{t}}{1} - \frac{t^2 e^t}{2}+ 2 \left(\frac{t^5 e^t}{120} \right)$

$\displaystyle y(t) = e^{t} + t e^{t} - \frac{1}{2} t^2 e^t+ \frac{1}{60} t^5 e^t$

Finalmente,

$\displaystyle \therefore y(t) = e^{t} + t e^{t} - \frac{1}{2} t^2 e^t+ \frac{1}{60} t^5 e^t$

Problema 5. Resolver $\displaystyle y'' + 9y = \cos{2t}$ si $y(0) = 1$ y $\displaystyle y \left(\frac{\pi}{2} \right) = -1$ .

Solución. Aplicando la transformada de Laplace en ambos miembros,

$\displaystyle y'' + 9y = \cos{2t}$

$\displaystyle \mathcal{L} [y'' + 9y] = \mathcal{L} [\cos{2t}]$

$\displaystyle \mathcal{L} [y''] + \mathcal{L} [9y] = \mathcal{L} [\cos{2t}]$

$\displaystyle \mathcal{L} [y''] + 9 \mathcal{L} [y] = \mathcal{L} [\cos{2t}]$

$\displaystyle s^2 Y(s) - s y(0) - y'(0) + 9 Y(s)= \frac{s}{s^2+4}$

Sabiendo que $y(0)=1$

$\displaystyle s^2 Y(s) - s (1) - y'(0) + 9 Y(s)= \frac{s}{s^2+4}$

$\displaystyle s^2 Y(s) - s - y'(0) + 9 Y(s)= \frac{s}{s^2+4}$

$\displaystyle (s^2 + 9) Y(s) - s - y'(0) = \frac{s}{s^2+4}$

Como no se tiene definido $y'(0)$ , sea $y'(0)=c$ .

$\displaystyle (s^2 + 9) Y(s) - s - c = \frac{s}{s^2+4}$

Continuando

$\displaystyle (s^2 + 9) Y(s) = s + c + \frac{s}{s^2+4}$

$\displaystyle Y(s) = \frac{s + c}{s^2+9} + \frac{s}{(s^2+4)(s^2+9)}$

$\displaystyle Y(s) = \frac{s}{s^2+9} + \frac{c}{s^2+9} + \frac{As+B}{s^2+4} + \frac{Cs+D}{s^2+9}$

$\displaystyle Y(s) = \frac{s}{s^2+9} + \frac{c}{s^2+9} + \frac{\frac{1}{5} s+0}{s^2+4} - \frac{\frac{1}{5} s+0}{s^2+9}$

$\displaystyle Y(s) = \frac{s}{s^2+9} + \frac{c}{s^2+9} + \frac{\frac{1}{5} s}{s^2+4} - \frac{\frac{1}{5} s}{s^2+9}$

$\displaystyle Y(s) = \frac{s}{s^2+9} + \frac{c}{s^2+9} + \frac{1}{5} \left(\frac{s}{s^2+4} \right) - \frac{1}{5} \left(\frac{s}{s^2+9} \right)$

$\displaystyle Y(s) = \frac{4}{5} \left(\frac{s}{s^2+9} \right) + \frac{c}{s^2+9} + \frac{1}{5} \left(\frac{s}{s^2+4} \right)$

Tomando $\mathcal{L}^{-1}$ en ambos lados

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[\frac{4}{5} \left(\frac{s}{s^2+9} \right) + \frac{c}{s^2+9} + \frac{1}{5} \left(\frac{s}{s^2+4} \right) \right]$

$\displaystyle \mathcal{L}^{-1} [Y(s)] = \frac{4}{5} \ \mathcal{L}^{-1} \left[\frac{s}{s^2+9} \right] + c \ \mathcal{L}^{-1} \left[\frac{1}{s^2+9} \right] + \frac{1}{5} \ \mathcal{L}^{-1} \left[\frac{s}{s^2+4} \right]$

$\displaystyle y(t) = \frac{4}{5} \cos{3t} + \frac{c}{3} \sin{3t} + \frac{1}{5} \cos{2t}$

Para determinar el valor $c$ , se toma $\displaystyle y \left(\frac{\pi}{2} \right) = -1$ .

$\displaystyle y \left( \frac{\pi}{2} \right) = \frac{4}{5} \cos{\left(3 \cdot \frac{\pi}{2} \right)} + \frac{c}{3} \sin{\left(3 \cdot \frac{\pi}{2} \right)} + \frac{1}{5} \cos{\left(2 \cdot \frac{\pi}{2} \right)}$

$\displaystyle y \left( \frac{\pi}{2} \right) = \frac{4}{5} \cos{\frac{3 \pi}{2}} + \frac{c}{3} \sin{\frac{3\pi}{2}} + \frac{1}{5} \cos{\pi}$

$\displaystyle -1 = \frac{4}{5} (0) + \frac{c}{3} (-1) + \frac{1}{5} (-1)$

$\displaystyle -1 = - \frac{c}{3} - \frac{1}{5}$

$\displaystyle \frac{c}{3} = 1 - \frac{1}{5}$

$\displaystyle \frac{c}{3} = \frac{4}{5}$

$\displaystyle c = \frac{12}{5}$

Ahora, sustituyendo el valor de $c$ en el resultado $y(t)$

$\displaystyle y(t) = \frac{4}{5} \cos{3t} + \frac{c}{3} \sin{3t} + \frac{1}{5} \cos{2t}$

$\displaystyle y(t) = \frac{4}{5} \cos{3t} + \frac{\frac{12}{5}}{3} \sin{3t} + \frac{1}{5} \cos{2t}$

$\displaystyle y(t) = \frac{4}{5} \cos{3t} + \frac{12}{15} \sin{3t} + \frac{1}{5} \cos{2t}$

$\displaystyle y(t) = \frac{4}{5} \cos{3t} + \frac{4}{5} \sin{3t} + \frac{1}{5} \cos{2t}$

$\displaystyle y(t) = \frac{4}{5} (\cos{3t} + \sin{3t}) + \frac{1}{5} \cos{2t}$

Finalmente,

$\displaystyle \therefore y(t) = \frac{4}{5} (\cos{3t} + \sin{3t}) + \frac{1}{5} \cos{2t}$

Solución de ecuaciones diferenciales ordinarias con coeficientes constantes utilizando la transformada de Laplace. Laplace.

Introducción

Problemas resueltos

Publicado por Cesar Reyes

Deja un comentario Cancelar la respuesta

Solución de ecuaciones diferenciales ordinarias con coeficientes constantes utilizando la transformada de Laplace. Laplace.

Introducción

Problemas resueltos

Comparte esto:

Relacionado

Publicado por Cesar Reyes

Deja un comentario Cancelar la respuesta