¿Cómo resolver una ecuación diferencial que involucra circuitos por el método de series de potencias, hallando sus constantes?
La ecuación diferencial a resolver es de la forma:
\[Lq''(t)+Rq'(t)+\frac{1}{C}q(t)=E(t)\]
Los valores de las constantes son los siguientes $L=\frac{1}{4}H$, $R(t)=\left(1+\frac{t}{8}\right)\Omega$, $C=4F$, $E(t)=0$, reemplazando dichos valores la ecuación diferencial a resolver es de la forma:
\[\frac{1}{4}q''(t)+\left(1+\frac{t}{8}\right)q'(t)+\frac{1}{4}q(t)=0\]
Se propone una solución en series de potencias y sus correspondientes derivadas de hasta segundo orden:
\[q(t)=\sum_{n=0}^{\infty}c_{n}t^{n}\]
\[q(t)'=\sum_{n=1}^{\infty}c_{n}nt^{n-1}\]
\[q(t)''=\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}\]
Reemplazando en nuestra ecuación diferencial tenemos:
\[\frac{1}{4}\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}+\left(1+\frac{t}{8}\right)\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{1}{4}\sum_{n=0}^{\infty}c_{n}t^{n}=0\]
Distribuyendo para la segunda sumatoria tenemos:
\[\frac{1}{4}\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}+\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{t}{8}\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{1}{4}\sum_{n=0}^{\infty}c_{n}t^{n}=0\]
Multiplicando el término $t$ que le aporta a la tercera sumatoria tendremos:
\[\frac{1}{4}\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}+\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{1}{8}\sum_{n=0}^{\infty}c_{n}nt^{n}+\frac{1}{4}\sum_{n=0}^{\infty}c_{n}t^{n}=0\]
Hacemos $k=n-2$ y $n=k+2$ para la primera sumatoria, $k=n-1$ y $n=k+1$ para la segunda sumatoria y $k=n$ para la tercera y cuarta sumatoria, reemplazamos y tenemos:
\[\frac{1}{4}\sum_{k=0}^{\infty}c_{k+2}(k+2)(k+1)t^{k}+\sum_{k=0}^{\infty}c_{k+1}(k+1)t^{k}+\frac{1}{8}\sum_{k=0}^{\infty}c_{k}kt^{k}+\frac{1}{4}\sum_{k=0}^{\infty}c_{k}t^{k}=0\]
Que se puede escribir de la forma:
\[\sum_{k=0}^{\infty}\left[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{8}c_{k}k+\frac{1}{4}c_{k}\right]t^{k}=0\]
Luego:
\[\sum_{k=0}^{\infty}\underbrace{\left[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{8}c_{k}k+\frac{1}{4}c_{k}\right]}_{=0}\underbrace{t^{k}}_{\neq0}=0\]
Al despejar la constante de mayor valor en términos de las constantes de valor tendremos:
\[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{8}c_{k}k+\frac{1}{4}c_{k}=0\]
\[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{4}c_{k}\left[\frac{k}{2}+1\right]=0\]
\[\frac{1}{4}c_{k+2}(k+2)(k+1)=-c_{k+1}(k+1)-\frac{1}{4}c_{k}\left[\frac{k}{2}+1\right]\]
\[c_{k+2}=-\frac{4(k+1)c_{k+1}}{(k+2)(k+1)}-\frac{4}{4}\frac{c_{k}}{(k+2)(k+1)}\left[\frac{k}{2}+1\right]\]
\[c_{k+2}=-\frac{4(k+1)c_{k+1}}{(k+2)(k+1)}-\frac{4}{4}\frac{c_{k}}{(k+2)(k+1)}\left[\frac{k+2}{2}\right]\]
\[c_{k+2}=-\frac{4c_{k+1}}{(k+2)}-\frac{c_{k}}{2(k+1)}\]
Le damos valores desde $k=0$ hasta el valor $k=3$ dado que se piden calcular los primeros $5$ términos de cada solución, por lo tanto:
Para $k=0$:
\[c_{0+2}=-\frac{4c_{0+1}}{(0+2)}-\frac{c_{0}}{2(0+1)}\]
\[c_{2}=-\frac{4c_{1}}{2}-\frac{c_{0}}{2(1)}\]
\[c_{2}=-\frac{4c_{1}}{2}-\frac{c_{0}}{2}\]
\[c_{2}=-2c_{1}-\frac{c_{0}}{2}\]
Para $k=1$:
\[c_{1+2}=-\frac{4c_{1+1}}{(1+2)}-\frac{c_{1}}{2(1+1)}\]
\[c_{3}=-\frac{4c_{2}}{3}-\frac{c_{1}}{2(2)}\]
\[c_{3}=-\frac{4c_{2}}{3}-\frac{c_{1}}{4}\]
\[c_{3}=-\frac{4}{3}\left(-2c_{1}-\frac{c_{0}}{2}\right)-\frac{c_{1}}{4}\]
\[c_{3}=\frac{8c_{1}}{3}+\frac{4c_{0}}{6}-\frac{c_{1}}{4}\]
\[c_{3}=\frac{8c_{1}}{3}-\frac{c_{1}}{4}+\frac{4c_{0}}{6}\]
\[c_{3}=\frac{32c_{1}}{12}-\frac{3c_{1}}{12}+\frac{4c_{0}}{6}\]
\[c_{3}=\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\]
Para $k=2$:
\[c_{2+2}=-\frac{4c_{2+1}}{(2+2)}-\frac{c_{2}}{2(2+1)}\]
\[c_{4}=-\frac{4c_{3}}{4}-\frac{c_{2}}{2(3)}\]
\[c_{4}=-\frac{4c_{3}}{4}-\frac{c_{2}}{6}\]
\[c_{4}=-\frac{4}{4}\left(\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\right)-\frac{1}{6}\left(-2c_{1}-\frac{c_{0}}{2}\right)\]
\[c_{4}=-\frac{29c_{1}}{12}-\frac{2c_{0}}{3}+\frac{4c_{1}}{12}+\frac{c_{0}}{12}\]
\[c_{4}=-\frac{29c_{1}}{12}-\frac{4c_{1}}{12}+\frac{2c_{0}}{3}+\frac{c_{0}}{12}\]
\[c_{4}=-\frac{29c_{1}}{12}-\frac{4c_{1}}{12}+\frac{8c_{0}}{12}+\frac{c_{0}}{12}\]
\[c_{4}=-\frac{33c_{1}}{12}+\frac{9c_{0}}{12}\]
\[c_{4}=-\frac{11c_{1}}{4}+\frac{3c_{0}}{4}\]
Para $k=3$:
\[c_{3+2}=-\frac{4c_{3+1}}{(3+2)}-\frac{c_{3}}{2(3+1)}\]
\[c_{5}=-\frac{4c_{4}}{5}-\frac{c_{3}}{2(4)}\]
\[c_{5}=-\frac{4c_{4}}{5}-\frac{c_{3}}{8}\]
\[c_{5}=-\frac{4}{5}\left(-\frac{11c_{1}}{4}+\frac{3c_{0}}{4}\right)-\frac{1}{6}\left(\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\right)\]
\[c_{5}=\frac{100c_{1}}{60}-\frac{12c_{0}}{20}-\frac{29c_{1}}{96}-\frac{4c_{0}}{42}\]
\[c_{5}=\frac{10c_{1}}{6}-\frac{6c_{0}}{10}-\frac{29c_{1}}{96}-\frac{2c_{0}}{21}\]
\[c_{5}=\frac{5c_{1}}{3}-\frac{3c_{0}}{5}-\frac{29c_{1}}{96}-\frac{2c_{0}}{21}\]
\[c_{5}=\frac{5c_{1}}{3}-\frac{29c_{1}}{96}-\frac{3c_{0}}{5}-\frac{2c_{0}}{21}\]
\[c_{5}=\frac{160c_{1}}{96}-\frac{29c_{1}}{96}-\frac{63c_{0}}{105}-\frac{10c_{0}}{105}\]
\[c_{5}=\frac{131c_{1}}{96}-\frac{39c_{0}}{105}\]
\[c_{5}=\frac{131c_{1}}{96}-\frac{13c_{0}}{35}\]
Luego la solución en serie de potencias corresponde a:
\[q(t)=c_{0}+c_{1}t+c_{2}t^{2}+c_{3}t^{3}+c_{4}t^{4}+c_{5}t^{5}+...\]
Reemplazamos las constantes y tendremos nuestra solución expresada en términos de $c_{0}$ y $c_{1}$:
\[q(t)=c_{0}+c_{1}t+\left(-2c_{1}-\frac{c_{0}}{2}\right)t^{2}+\left(\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\right)t^{3}+\left(-\frac{11c_{1}}{4}+\frac{3c_{0}}{4}\right)t^{4}+\left(\frac{131c_{1}}{96}-\frac{13c_{0}}{35}\right)t^{5}+...\]
Factorizamos en terminos de $c_{0}$ y $c_{1}$ y tendremos:
\[q(t)=c_{0}\left(1-\frac{t^{2}}{2}+\frac{2t^{3}}{3}+\frac{3t^{4}}{4}-\frac{13t^{5}}{35}+...\right)+c_{1}\left(t-2t^{2}+\frac{29t^{3}}{12}-\frac{11t^{4}}{4}+\frac{131t^{5}}{96}+...\right)\]
Aplicando condiciones iniciales en $q(0)=2$ tendremos:
\[q(0)=c_{0}\left(1-\frac{(0)^{2}}{2}+\frac{2(0)^{3}}{3}+\frac{3(0)^{4}}{4}-\frac{13(0)^{5}}{35}+...\right)+c_{1}\left((0)-2(0)^{2}+\frac{29(0)^{3}}{12}-\frac{11(0)^{4}}{4}+\frac{131(0)^{5}}{96}+...\right)\]
\[q(0)=2=c_{0}(1)+c_{1}(0)\]
Ahora cuando $\frac{dq(0)}{dt}=0$ lo que equivale a derivar la expresión anterior:
\[\frac{dq(t)}{dt}=2\left(-t-\frac{3t^{2}}{3}+\frac{12t^{3}}{4}+\frac{215t^{4}}{105}+...\right)+c_{1}\left(1-4t+\frac{87t^{2}}{12}-\frac{44t^{3}}{4}+\frac{655t^{4}}{96}+...\right)\]
\[\frac{dq(t)}{dt}=0=2\left(-(0)-\frac{3(0)^{2}}{3}+\frac{12(0)^{3}}{4}+\frac{215(0)^{4}}{105}+...\right)+c_{1}\left(1-4(0)+\frac{87(0)^{2}}{12}-\frac{44(0)^{3}}{4}+\frac{655(0)^{4}}{96}+...\right)\]
\[0=2c_{1}(1)\]
\[0=c_{1}\]
Luego la correspondiente solución a mi ecuación diferencial es:
\[q(t)=2\left(1-\frac{t^{2}}{2}+\frac{2t^{3}}{3}+\frac{3t^{4}}{4}-\frac{13t^{5}}{35}+...\right)\]
De acuerdo a las condiciones iniciales dadas.
\[Lq''(t)+Rq'(t)+\frac{1}{C}q(t)=E(t)\]
Los valores de las constantes son los siguientes $L=\frac{1}{4}H$, $R(t)=\left(1+\frac{t}{8}\right)\Omega$, $C=4F$, $E(t)=0$, reemplazando dichos valores la ecuación diferencial a resolver es de la forma:
\[\frac{1}{4}q''(t)+\left(1+\frac{t}{8}\right)q'(t)+\frac{1}{4}q(t)=0\]
Se propone una solución en series de potencias y sus correspondientes derivadas de hasta segundo orden:
\[q(t)=\sum_{n=0}^{\infty}c_{n}t^{n}\]
\[q(t)'=\sum_{n=1}^{\infty}c_{n}nt^{n-1}\]
\[q(t)''=\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}\]
Reemplazando en nuestra ecuación diferencial tenemos:
\[\frac{1}{4}\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}+\left(1+\frac{t}{8}\right)\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{1}{4}\sum_{n=0}^{\infty}c_{n}t^{n}=0\]
Distribuyendo para la segunda sumatoria tenemos:
\[\frac{1}{4}\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}+\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{t}{8}\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{1}{4}\sum_{n=0}^{\infty}c_{n}t^{n}=0\]
Multiplicando el término $t$ que le aporta a la tercera sumatoria tendremos:
\[\frac{1}{4}\sum_{n=2}^{\infty}c_{n}n(n-1)t^{n-2}+\sum_{n=1}^{\infty}c_{n}nt^{n-1}+\frac{1}{8}\sum_{n=0}^{\infty}c_{n}nt^{n}+\frac{1}{4}\sum_{n=0}^{\infty}c_{n}t^{n}=0\]
Hacemos $k=n-2$ y $n=k+2$ para la primera sumatoria, $k=n-1$ y $n=k+1$ para la segunda sumatoria y $k=n$ para la tercera y cuarta sumatoria, reemplazamos y tenemos:
\[\frac{1}{4}\sum_{k=0}^{\infty}c_{k+2}(k+2)(k+1)t^{k}+\sum_{k=0}^{\infty}c_{k+1}(k+1)t^{k}+\frac{1}{8}\sum_{k=0}^{\infty}c_{k}kt^{k}+\frac{1}{4}\sum_{k=0}^{\infty}c_{k}t^{k}=0\]
Que se puede escribir de la forma:
\[\sum_{k=0}^{\infty}\left[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{8}c_{k}k+\frac{1}{4}c_{k}\right]t^{k}=0\]
Luego:
\[\sum_{k=0}^{\infty}\underbrace{\left[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{8}c_{k}k+\frac{1}{4}c_{k}\right]}_{=0}\underbrace{t^{k}}_{\neq0}=0\]
Al despejar la constante de mayor valor en términos de las constantes de valor tendremos:
\[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{8}c_{k}k+\frac{1}{4}c_{k}=0\]
\[\frac{1}{4}c_{k+2}(k+2)(k+1)+c_{k+1}(k+1)+\frac{1}{4}c_{k}\left[\frac{k}{2}+1\right]=0\]
\[\frac{1}{4}c_{k+2}(k+2)(k+1)=-c_{k+1}(k+1)-\frac{1}{4}c_{k}\left[\frac{k}{2}+1\right]\]
\[c_{k+2}=-\frac{4(k+1)c_{k+1}}{(k+2)(k+1)}-\frac{4}{4}\frac{c_{k}}{(k+2)(k+1)}\left[\frac{k}{2}+1\right]\]
\[c_{k+2}=-\frac{4(k+1)c_{k+1}}{(k+2)(k+1)}-\frac{4}{4}\frac{c_{k}}{(k+2)(k+1)}\left[\frac{k+2}{2}\right]\]
\[c_{k+2}=-\frac{4c_{k+1}}{(k+2)}-\frac{c_{k}}{2(k+1)}\]
Le damos valores desde $k=0$ hasta el valor $k=3$ dado que se piden calcular los primeros $5$ términos de cada solución, por lo tanto:
Para $k=0$:
\[c_{0+2}=-\frac{4c_{0+1}}{(0+2)}-\frac{c_{0}}{2(0+1)}\]
\[c_{2}=-\frac{4c_{1}}{2}-\frac{c_{0}}{2(1)}\]
\[c_{2}=-\frac{4c_{1}}{2}-\frac{c_{0}}{2}\]
\[c_{2}=-2c_{1}-\frac{c_{0}}{2}\]
Para $k=1$:
\[c_{1+2}=-\frac{4c_{1+1}}{(1+2)}-\frac{c_{1}}{2(1+1)}\]
\[c_{3}=-\frac{4c_{2}}{3}-\frac{c_{1}}{2(2)}\]
\[c_{3}=-\frac{4c_{2}}{3}-\frac{c_{1}}{4}\]
\[c_{3}=-\frac{4}{3}\left(-2c_{1}-\frac{c_{0}}{2}\right)-\frac{c_{1}}{4}\]
\[c_{3}=\frac{8c_{1}}{3}+\frac{4c_{0}}{6}-\frac{c_{1}}{4}\]
\[c_{3}=\frac{8c_{1}}{3}-\frac{c_{1}}{4}+\frac{4c_{0}}{6}\]
\[c_{3}=\frac{32c_{1}}{12}-\frac{3c_{1}}{12}+\frac{4c_{0}}{6}\]
\[c_{3}=\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\]
Para $k=2$:
\[c_{2+2}=-\frac{4c_{2+1}}{(2+2)}-\frac{c_{2}}{2(2+1)}\]
\[c_{4}=-\frac{4c_{3}}{4}-\frac{c_{2}}{2(3)}\]
\[c_{4}=-\frac{4c_{3}}{4}-\frac{c_{2}}{6}\]
\[c_{4}=-\frac{4}{4}\left(\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\right)-\frac{1}{6}\left(-2c_{1}-\frac{c_{0}}{2}\right)\]
\[c_{4}=-\frac{29c_{1}}{12}-\frac{2c_{0}}{3}+\frac{4c_{1}}{12}+\frac{c_{0}}{12}\]
\[c_{4}=-\frac{29c_{1}}{12}-\frac{4c_{1}}{12}+\frac{2c_{0}}{3}+\frac{c_{0}}{12}\]
\[c_{4}=-\frac{29c_{1}}{12}-\frac{4c_{1}}{12}+\frac{8c_{0}}{12}+\frac{c_{0}}{12}\]
\[c_{4}=-\frac{33c_{1}}{12}+\frac{9c_{0}}{12}\]
\[c_{4}=-\frac{11c_{1}}{4}+\frac{3c_{0}}{4}\]
Para $k=3$:
\[c_{3+2}=-\frac{4c_{3+1}}{(3+2)}-\frac{c_{3}}{2(3+1)}\]
\[c_{5}=-\frac{4c_{4}}{5}-\frac{c_{3}}{2(4)}\]
\[c_{5}=-\frac{4c_{4}}{5}-\frac{c_{3}}{8}\]
\[c_{5}=-\frac{4}{5}\left(-\frac{11c_{1}}{4}+\frac{3c_{0}}{4}\right)-\frac{1}{6}\left(\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\right)\]
\[c_{5}=\frac{100c_{1}}{60}-\frac{12c_{0}}{20}-\frac{29c_{1}}{96}-\frac{4c_{0}}{42}\]
\[c_{5}=\frac{10c_{1}}{6}-\frac{6c_{0}}{10}-\frac{29c_{1}}{96}-\frac{2c_{0}}{21}\]
\[c_{5}=\frac{5c_{1}}{3}-\frac{3c_{0}}{5}-\frac{29c_{1}}{96}-\frac{2c_{0}}{21}\]
\[c_{5}=\frac{5c_{1}}{3}-\frac{29c_{1}}{96}-\frac{3c_{0}}{5}-\frac{2c_{0}}{21}\]
\[c_{5}=\frac{160c_{1}}{96}-\frac{29c_{1}}{96}-\frac{63c_{0}}{105}-\frac{10c_{0}}{105}\]
\[c_{5}=\frac{131c_{1}}{96}-\frac{39c_{0}}{105}\]
\[c_{5}=\frac{131c_{1}}{96}-\frac{13c_{0}}{35}\]
Luego la solución en serie de potencias corresponde a:
\[q(t)=c_{0}+c_{1}t+c_{2}t^{2}+c_{3}t^{3}+c_{4}t^{4}+c_{5}t^{5}+...\]
Reemplazamos las constantes y tendremos nuestra solución expresada en términos de $c_{0}$ y $c_{1}$:
\[q(t)=c_{0}+c_{1}t+\left(-2c_{1}-\frac{c_{0}}{2}\right)t^{2}+\left(\frac{29c_{1}}{12}+\frac{2c_{0}}{3}\right)t^{3}+\left(-\frac{11c_{1}}{4}+\frac{3c_{0}}{4}\right)t^{4}+\left(\frac{131c_{1}}{96}-\frac{13c_{0}}{35}\right)t^{5}+...\]
Factorizamos en terminos de $c_{0}$ y $c_{1}$ y tendremos:
\[q(t)=c_{0}\left(1-\frac{t^{2}}{2}+\frac{2t^{3}}{3}+\frac{3t^{4}}{4}-\frac{13t^{5}}{35}+...\right)+c_{1}\left(t-2t^{2}+\frac{29t^{3}}{12}-\frac{11t^{4}}{4}+\frac{131t^{5}}{96}+...\right)\]
Aplicando condiciones iniciales en $q(0)=2$ tendremos:
\[q(0)=c_{0}\left(1-\frac{(0)^{2}}{2}+\frac{2(0)^{3}}{3}+\frac{3(0)^{4}}{4}-\frac{13(0)^{5}}{35}+...\right)+c_{1}\left((0)-2(0)^{2}+\frac{29(0)^{3}}{12}-\frac{11(0)^{4}}{4}+\frac{131(0)^{5}}{96}+...\right)\]
\[q(0)=2=c_{0}(1)+c_{1}(0)\]
Ahora cuando $\frac{dq(0)}{dt}=0$ lo que equivale a derivar la expresión anterior:
\[\frac{dq(t)}{dt}=2\left(-t-\frac{3t^{2}}{3}+\frac{12t^{3}}{4}+\frac{215t^{4}}{105}+...\right)+c_{1}\left(1-4t+\frac{87t^{2}}{12}-\frac{44t^{3}}{4}+\frac{655t^{4}}{96}+...\right)\]
\[\frac{dq(t)}{dt}=0=2\left(-(0)-\frac{3(0)^{2}}{3}+\frac{12(0)^{3}}{4}+\frac{215(0)^{4}}{105}+...\right)+c_{1}\left(1-4(0)+\frac{87(0)^{2}}{12}-\frac{44(0)^{3}}{4}+\frac{655(0)^{4}}{96}+...\right)\]
\[0=2c_{1}(1)\]
\[0=c_{1}\]
Luego la correspondiente solución a mi ecuación diferencial es:
\[q(t)=2\left(1-\frac{t^{2}}{2}+\frac{2t^{3}}{3}+\frac{3t^{4}}{4}-\frac{13t^{5}}{35}+...\right)\]
De acuerdo a las condiciones iniciales dadas.
Comentarios
Publicar un comentario