¿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″
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″
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.
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