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http://sandymillin.wordpress.com/iateflwebinar2024
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For more information, visit-www.vavaclasses.com
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1. Programa Nacional de Formación en Sistema de Calidad y Ambiente
Ecuaciones Lineales de Orden Superior
Participante:
María Rivas CI.: 25.546.373
Sección: SC 4100
Barquisimeto, 2021
2. Ecuaciones Lineales de Orden Superior
Ecuaciones Homogéneas:
1. Resolver 4𝑦” + 𝑦´ = 0
Solución: Al examinar la ecuación la completamos de la siguiente manera:
4𝑦” + 𝑦 + 0𝑦 = 0
Ahora llevamos la ecuación 4𝑦” + 𝑦 + 0𝑦 = 0 a 4𝑚2 + 𝑚 + 0 = 0, la cual es el polinomio
característico, luego sacamos las raíces del polinomio 4𝑚2+ 𝑚 = 0, primero sacamos factor
común de la siguiente manera 4𝑚2 + 𝑚=4𝑚 (𝑚+¼) de esta manera obtenemos que las
raíces son 𝑚1 = 0 y 𝑚2 = −¼ , las cuales son raíces reales y distintas.
Sabemos que 𝑦 = c1𝑒 m1x+ c2𝑒 m2x+…+ cn𝑒 mnx, es la solución general para este tipo de
ecuación, por lo que adecuamos la ecuación a la solución general nos queda de la siguiente
manera 𝑦 = c1𝑒 0x + c2𝑒− ¼ ; donde 𝑚1 = 0 y 𝑚2 = −¼
Así la solución al problema es: 𝑦 = c1 + c2𝑒− ¼
2.- Resolver
𝑦” - 36 𝑦 = 0
Solución: Completamos la ecuación 𝑦” + 0 - 36𝑦 = 0, ahora bien llevamos la ecuación al
polinomio característico 𝑚2 + 0 - 36 = 0, factorizamos con la siguiente formula a2-b2=(a+ b).
(a-b) pero arreglamos la ecuación y nos queda que: 𝑚2-62 así obtenemos las siguientes raíces
𝑚1= 6 ; 𝑚2 = -6 ; del polinomio vemos que ambos son raíces reales y distintas.
Sabemos que por teoría 𝑦 = c1𝑒 m1x+ c2𝑒 m2x+…+ cn𝑒 mnx, es la solución general para
resolver este ejercicio.
Adecuamos la solución general al ejercicio con las raíces obtenidas 𝑦 = c1𝑒6𝑥 + c2𝑒−6𝑥
así tenemos que la solución es: 𝑦 = c1𝑒6x + c2𝑒-6x
Así tenemos que la solución es: 𝑦 = c1𝑒6x + c2𝑒-6x
Ecuaciones Lineales:
1.- Veamos si 𝑦 = 3𝑒2x + 𝑒-2x -3𝑥 , Es solución del problema de valores iniciales
𝑦” - 4𝑦 = 12𝑥 ; 𝑦 0 = 0 𝑦´(0) = 1
3. Solución: Comprobamos que 𝑦 = 3𝑒2x + 𝑒-2x -3𝑥, es una solución para
𝑦” - 4𝑦 = 12𝑥, luego derivamos 2 veces a 𝑦 = 3𝑒2x + 𝑒-2x -3𝑥
𝑦´ = 6𝑒2x - 2𝑒-2x - 3
𝑦” = 12𝑒2x + 4𝑒-2x
Ahora sustituimos las derivadas obtenidas anteriormente en 𝑦” - 4𝑦 = 12𝑥
12𝑒2x + 4𝑒-2x-4(3𝑒2x + 𝑒-2x -3𝑥) = 12𝑥
12𝑒2x + 4𝑒-2x - 12𝑒2x + 4𝑒-2x + 12𝑥 = 12𝑥
12𝑥 = 12𝑥
Por lo que 𝑦 =3𝑒2x + 𝑒-2x , es efectivamente una solución de
𝑦” - 4𝑦 = 12𝑥
Comprobamos si las condiciones iniciales
𝑦 0 = 4 ; 𝑦´(0) = 1 , también cumplen con nuestra solución, el primer parámetro 𝑦 0 = 4,
sabiendo que cero(0) lo sustituimos por 𝑥 y a 4 por 𝑦, en: 𝑦 = 3𝑒2x + 𝑒-2x -3𝑥
Entonces: 4 = 3𝑒2 (0) + 𝑒-2 (0)- 3 .(0)
4 = 3𝑒0 + 𝑒-0 _ 3 .(0) ; Sabiendo que 𝑒0= 1
4 = 3 . (1) + 1
4 = 3+1
4 = 4
Hacemos lo mismo con el segundo para sustituimos cero (0), en donde va 𝑥 y a 1 donde va
𝑦´, pero con la primera derivada por la condición del parámetro 𝑦´ = 6𝑒2x - 2 𝑒-2x - 3
1 = 6𝑒2 (0) - 2𝑒-2 (0)- 3
1 = 6𝑒0 - 2 𝑒-0 _ 3 ; Sabiendo que 𝑒0= 1
1 = 6 – 2 – 3
1 = 1
La ecuación diferencial es lineal, los coeficientes 𝑦 g(𝑥) y a2(𝑥) = 1≠ 0 en todo intervalo I
que contenga a 𝑥 = 0. Según el teorema: Sean an(𝑥), an-1(𝑥),…,a1 𝑥 ,a0(𝑥) y g(𝑥) continuas
en un intervalo I, y sea an(𝑥)≠0 para toda (𝑥) del intervalo I. Si (𝑥)= (𝑥0) es cualquier punto
en el intervalo, existe una solución en dicho intervalo 𝑦(𝑥) del problema de valores iniciales
4. representado por las ecuaciones (I), que es única, debemos concluir que la función dada es
la única solución en (I).
2.- Resolvemos: 𝑦 = c𝑥2 + 𝑥 + 3, es solución del problema de valores iniciales
𝑥2𝑦”- 2𝑥𝑦´ + 2𝑦 = 6 ; 𝑦(0) = 3 ; 𝑦´(0) = 1
Solución: Comprobamos que 𝑦 = c𝑥 2+ 𝑥 + 3, es una solución para 𝑥2𝑦”- 2𝑥𝑦´ + 2𝑦 = 6,
derivamos 2 veces a 𝑦 = c𝑥 2+ 𝑥 + 3
𝑦´= 2𝑐𝑥 + 1 + 0
𝑦” = 2𝑐
Ahora sustituimos las derivadas obtenidas anteriormente en 𝑥2𝑦”- 2𝑥𝑦´ + 2𝑦 = 6
𝑥2 (2𝑐) - 2𝑥(2𝑐𝑥 + 1) + 2(c𝑥2 + 𝑥 + 3) = 6
2c𝑥 2- 4c𝑥2 - 2𝑥 + 2c𝑥2 + 2𝑥 + 6 = 6
6 = 6
Por lo que 𝑦 = c𝑥2 + 𝑥 + 3, es efectivamente una solución para 𝑥 2 𝑦”- 2𝑥𝑦´ + 2𝑦 = 6,
entonces comprobamos las condiciones iniciales 𝑦(0) = 3 , 𝑦´(0) = 1, también
cumplen con nuestra solución.
El primer parámetro 𝑦(0) = 3, sabiendo que 𝑥 = 0 y 𝑦 = 3
𝑦 = c𝑥2 + 𝑥 + 3
3 = c(0) + (0) + 3
3 = 3
Para el segundo parámetro 𝑦´(0) = 1, hacemos 𝑥 = 0 y 𝑦´= 1(la derivada) en 𝑦´= 2𝑐𝑥
+ 1
1 = 2c(0) + 1
1 = 1
NO existe solución única para el problema, aunque satisface la mayor parte de las
condiciones del teorema Sean an(𝑥), an-1(𝑥),…,a1 𝑥 ,a0(𝑥) y g(𝑥) continuas en un
intervalo I, y sea an(𝑥)≠0 para toda (𝑥) del intervalo I. Si (𝑥)= (𝑥0) es cualquier punto
en el intervalo, existe una solución en dicho intervalo 𝑦(𝑥) del problema de valores
iniciales representado por las ecuaciones (I), que es única, debemos concluir que la
función dada es la única solución en (I).
Las dificultades obvias están en que a22 𝑥 = 𝑥2, es cero cuando 𝑥 = 0