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**PROYECTO DE ECUACIONES DIFERENCIALES**

**INTEGRANTES:**

*JOHANNA ERMELINDA OSORIO RUIZ CARNET OR12017*

*JOSE LUIS VILLALOBOS MARTINEZ CARNET VM12038*

PROYECTO DE ECUACIONES DIFERENCIALES

INTEGRANTES:

JOHANNA ERMELINDA OSORIO RUIZ CARNET OR12017

JOSE LUIS VILLALOBOS MARTINEZ CARNET VM12038

Problema 1: Resolver la ecuación diferencial con SageMath. Consejo: Mira documentación en el libro de texto. Por su propia ecuación diferencial personal, sean P y Q dos dígitos distintos de cero en el número de identificación de estudiante, y consideran la ecuación diferencial.

dy/dx=cos(x−qy)/pdy/dx= cos(x-qy)/p

Encontrar una solución general simbólica mediante un sistema de álgebra computacional y / o alguna combinación de las técnicas que se enumeran en este capítulo (Edwards y Penney

reset()
a,b,x,C,z=var('a,b,x,C,z')
y = function('y')(x)
solution = desolve(diff(y,x) == (1/11)*cos(x-14*y) , y,ivar=x,contrib_ode=True,show_method=True)
mySolution=solution[0][0]
show(mySolution)
−1115 3log⁡(−3cos⁡(−x+14 y(x))+3+5 sin⁡(−x+14 y(x))3cos⁡(−x+14 y(x))+3−5 sin⁡(−x+14 y(x)))=C\displaystyle -\frac{11}{15} \, \sqrt{3} \log\left(-\frac{\sqrt{3} \cos\left(-x + 14 \, y\left(x\right)\right) + \sqrt{3} + 5 \, \sin\left(-x + 14 \, y\left(x\right)\right)}{\sqrt{3} \cos\left(-x + 14 \, y\left(x\right)\right) + \sqrt{3} - 5 \, \sin\left(-x + 14 \, y\left(x\right)\right)}\right) = C

b) Determinar la solución particular simbólica correspondiente con varias condiciones iniciales típicas de la forma y(x_0)=y_0. lo que haremos es encontrar valores diferentes para C para que se ajuste para diferentes condiciones iniciales.

Aplicar las condiciones iniciales x0 = 1, y0 = 1


ss=mySolution.lhs()
C=(ss).substitute({y(x):1}).substitute(x=1)
show(C)
equation1=ss.substitute({y(x):z})==1
equation2=-15/(11*sqrt(3))*equation1
show(equation2)
−1115 3log⁡(−3cos⁡(13)+3+5 sin⁡(13)3cos⁡(13)+3−5 sin⁡(13))\displaystyle -\frac{11}{15} \, \sqrt{3} \log\left(-\frac{\sqrt{3} \cos\left(13\right) + \sqrt{3} + 5 \, \sin\left(13\right)}{\sqrt{3} \cos\left(13\right) + \sqrt{3} - 5 \, \sin\left(13\right)}\right)
log⁡(−3cos⁡(−x+14 z)+3+5 sin⁡(−x+14 z)3cos⁡(−x+14 z)+3−5 sin⁡(−x+14 z))=−511 3\displaystyle \log\left(-\frac{\sqrt{3} \cos\left(-x + 14 \, z\right) + \sqrt{3} + 5 \, \sin\left(-x + 14 \, z\right)}{\sqrt{3} \cos\left(-x + 14 \, z\right) + \sqrt{3} - 5 \, \sin\left(-x + 14 \, z\right)}\right) = -\frac{5}{11} \, \sqrt{3}

le aplicamos las condiciones iniciales de nuevo x0 = 0, y0 = 0


mySolution= solution[0][0]
ss1=mySolution.lhs()
C1=(ss1).substitute({y(x):0}).substitute(x=0)
show(C1)
−1115i 3π\displaystyle -\frac{11}{15} i \, \sqrt{3} \pi

la aplicación de las condiciones iniciales de nuevo x0 = 10, y0 = 50


ss2=mySolution.lhs()
C2=(ss2).substitute({y(x):50}).substitute(x=10)
show(C2)
x,C,z=var('x,C,z')
y = function('y')(x)
sol = desolve(diff(y,x) == (1/11)*cos(x-14*y) , y,ivar=x,contrib_ode=True,show_method=True)
mySolution=solution[0][0]
ss=mySolution.lhs()
equation1=ss.substitute({y(x):z})==1
equation2=-15/(11*sqrt(3))*equation1
solve(equation2,z)
implicit_plot(equation2,(x,-5,5),(z,-5,5))
−1115 3log⁡(−3cos⁡(690)+3+5 sin⁡(690)3cos⁡(690)+3−5 sin⁡(690))\displaystyle -\frac{11}{15} \, \sqrt{3} \log\left(-\frac{\sqrt{3} \cos\left(690\right) + \sqrt{3} + 5 \, \sin\left(690\right)}{\sqrt{3} \cos\left(690\right) + \sqrt{3} - 5 \, \sin\left(690\right)}\right)
[] 


%md
c) Defina los posibles valores de a y b tales que la línea recta y=a*x+b sea una curva solución de la ecuación.
En La ecuación diferencial,sustituyendo tenemos que:



x,C,z=var('x,C,z')
y=function('y')(x)
sol=desolve(diff(y,x) == (1/11)*cos(x-14*y) , y,ivar=x,contrib_ode=True,show_method=True)
mySolution=solution[0][0]
ss=mySolution.lhs()
equation1=ss.substitute({y(x):z})==1
equation2=-15/(11*sqrt(3))*equation1
solve(equation2,z)
implicit_plot(equation2,(x,-5,5),(z,-5,5))
[] 


a,b,x,C,z=var('a,b,x,C,z')
y = function('y')(x)
equation3= diff(y,x)== (1/11)*cos(x-14*y)
k1=(equation3).substitute({y:(a*x+b)}).substitute({diff(y,x):a})
show(k1)
#factorizando,llegamos a que:
equation4= a== (1/11)*cos((1-14*a)*x - 14*b)
show(equation4)

#a continuacion,haciendo 14*a-1=0 se llega a que:
equation5= 14*a -1==0
solution1=solve(equation5,a)
mySolution1= solution1[0]
show(mySolution1)

#entonces, ahora puedes resolver a b,sustituyendo el valor de una, por lo que deberia de ser:
equation6=show(equation4.substitute ({a:1/14}))
solution2=solve(cos(-14*b)-11/14, b)
mySolution2=solution2[0]
show(mySolution2)

#Por lo tanto la solución es:
equation7= y == a*x+b
show(equation7)
final=equation7.substitute({a:mySolution1}).substitute({b:mySolution2})
show(final)
a=111 cos⁡(−14 ax−14 b+x)\displaystyle a = \frac{1}{11} \, \cos\left(-14 \, a x - 14 \, b + x\right)
a=111 cos⁡(−(14 a−1)x−14 b)\displaystyle a = \frac{1}{11} \, \cos\left(-{\left(14 \, a - 1\right)} x - 14 \, b\right)
a=(114)\displaystyle a = \left(\frac{1}{14}\right)
(114)=111 cos⁡(−14 b)\displaystyle \left(\frac{1}{14}\right) = \frac{1}{11} \, \cos\left(-14 \, b\right)
b=114 arccos⁡(1114)\displaystyle b = \frac{1}{14} \, \arccos\left(\frac{11}{14}\right)
y(x)=ax+b\displaystyle y\left(x\right) = a x + b
y(x)=x(a=(114))+(b=114 arccos⁡(1114))\displaystyle y\left(x\right) = x (a = \left(\frac{1}{14}\right)) + (b = \frac{1}{14} \, \arccos\left(\frac{11}{14}\right))
#d) Grafiqué un campo direccional y algunas curvas solución. ¿Puede establecer alguna conexión entre la solución simbólica y sus curvas solución.

from sage.calculus.desolvers import desolve_rk4
x,y=var('x,y')
slopefield=plot_slope_field((1/11)*cos(x-14*y),(x,-5,4),(y,-5,4))
#una instancia de una variable f
f=plot([])       #crear instancias de f como una trama vacía
#loop over to get many plots
#ajustes dinámicos,aviso de color RGB
for k in [1, 2, 3, 4, 5, 5.5, 8, 9, 10, 11, 11.3, 11.5, 11.7, 12]:
    f=f+line(desolve_rk4((1/11)*cos(x-14*y),y,ics=[k/2-3,-k/2+3],end_points=[-5,5],step=0.01), rgbcolor=(k/12,0,1-k/12))
#Mostrar todas las curvas y campos de pendientes
(slopefield+f).show(xmin=-5,xmax=5,ymin=-5,ymax=5)


from sage.calculus.desolvers import desolve_rk4
x,y=var('x,y')
slopefield=plot_slope_field((1/1)*cos(x-14*y),(x,-5,5),(y,-5,5))
#una instancia de una variable f
f=plot([])     #crear instancias de f como una trama vacía
#puntos por encima de la trama
pts=points([(-2.5,2.5),(-1,-2),(-2,1),(-1,2),(2,0), (1,0.5),(1,-1)],size=80,color='red')
#loop over to get many plots
#ajustes dinámicos aviso de color RGB
for ics in [[-2.5,2.5],[-1,-2],[-2,1],[-1,2],[2,0],[1,0.5],[1,-1]]:
    f=f+line(desolve_rk4((1/11)*cos(x-14*y),y,ics,end_points=[-5,5],step=0.05))
#mostrar todos los campos curvas + pendiente
(slopefield+pts+f).show(xmin=-5,xmax=5,ymin=-5,ymax=5)

problema2: #encontrar la solución de la ecuacion (D-2)3(D2+9)y=x2ex+xsin(3x); con valores iniciales y(0)=y'(0)=y"(0)=y3(0)=y4(0)=1


# Utilizo la ecuación caracteristica para encontrar las raices de la ecuación diferencial
r=var('r')
solve([(r-2)^3*(r^2+9)==0],r)
# Escribo la solución complementaria
var('x,c1,c2,c3,c4,c5')
yc= function('y')(x)
[r == (-3*I), r == (3*I), r == 2] (x, c1, c2, c3, c4, c5) 
# en c2 y c3 los multiplicó por x y x^2 respectivamente por que hay raices repetidas
yc(x)= c1*exp(2*x)+c2*(x)*exp(2*x)+c3*(x^2)*exp(2*x)+c4*cos(3*x)+c5*sin(3*x)
# Ahora para la solución particular tengo
# Utilizó la parte no homogenea de la ecuación diferencial
f= function ('y')(x)
f(x)= x^2*exp(x)+x*sin(3*x)


# Escribiendo la solución trivial para f(x), recordar que L[y]=f(x)
var('A,B,C,D,E,F,G')
yp= function('y')(x)
(A, B, C, D, E, F, G) 
# Multiplicó por x para eliminar multiplicidad de la solución de ensayo
yp(x)= A*exp(x)+B*(x)*exp(x)+C*(x^2)*exp(x)+x*((D*x+E)*cos(3*x)+(F*x+G)*sin(3*x))
# Ahora trabajo con yp para encontrar sus coeficientes
k=function('k')(x)
D1= diff(yp,x,2)+9*yp
D2= diff(D1,x,1)-2*D1
D3= diff(D2,x,1)-2*D2
D4= diff(D3,x,1)-2*D3
k= D4


# Resuelvo para poder hallar las constantes de la solución de ensayo tomando valores al azar
solve([k(0)==f(0),k(1)==f(1),k(6)==f(6),k(2)==f(2),k(5)==f(5),k(3)==f(3),k(4)==f(4)],A,B,C,D,E,F,G)
[[A == (-267/250), B == (-14/25), C == (-1/10), D == (-23/13182), E == (-251/114244), F == (-3/8788), G == (1379/514098)]] 
# Reescribimos la solución de ensayo con los valores encontrados
yp(x)= (-267/250)*exp(x)+(-14/25)*x*exp(x)+(-1/10)*(x^2)*exp(x)+x*(((-23/13182)*x+(-251/114244))*cos(3*x)+((-3/8788)*x+(1379/514098))*sin(3*x))
show(yp)
x ↦ −110 x2ex−11028196 (3 (598 x+753)cos⁡(3 x)+(351 x−2758)sin⁡(3 x))x−1425 xex−267250 ex\displaystyle x \ {\mapsto}\ -\frac{1}{10} \, x^{2} e^{x} - \frac{1}{1028196} \, {\left(3 \, {\left(598 \, x + 753\right)} \cos\left(3 \, x\right) + {\left(351 \, x - 2758\right)} \sin\left(3 \, x\right)\right)} x - \frac{14}{25} \, x e^{x} - \frac{267}{250} \, e^{x}
# Ahora escribo la solución general de la ecuación diferencial y encuentro las constantes de la solución complementaria
y= function('y')(x)
y(x)= yc(x)+ yp(x)


# Utilizó show para que la expresión se vea mejor
show(y)
solve([y(0)==1,diff(y,x,1)(0)==1,diff(y,x,2)(0)==1,diff(y,x,3)(0)==1,diff(y,x,4)(0)==1],c1,c2,c3,c4,c5)
x ↦ c3x2e(2 x)+c2xe(2 x)−110 x2ex−11028196 (3 (598 x+753)cos⁡(3 x)+(351 x−2758)sin⁡(3 x))x+c4cos⁡(3 x)+c1e(2 x)−1425 xex+c5sin⁡(3 x)−267250 ex\displaystyle x \ {\mapsto}\ c_{3} x^{2} e^{\left(2 \, x\right)} + c_{2} x e^{\left(2 \, x\right)} - \frac{1}{10} \, x^{2} e^{x} - \frac{1}{1028196} \, {\left(3 \, {\left(598 \, x + 753\right)} \cos\left(3 \, x\right) + {\left(351 \, x - 2758\right)} \sin\left(3 \, x\right)\right)} x + c_{4} \cos\left(3 \, x\right) + c_{1} e^{\left(2 \, x\right)} - \frac{14}{25} \, x e^{x} + c_{5} \sin\left(3 \, x\right) - \frac{267}{250} \, e^{x}
[[c1 == (774782/371293), c2 == (-43330/28561), c3 == (1020/2197), c4 == (-1737019/92823250), c5 == (-4850123/556939500)]] 
# Reescribimos la solución general con los valores encontrados de las constantes de la solución complementaria
y(x)= (774782/371293)*exp(2*x)+(-43330/28561)*(x)*exp(2*x)+(1020/2197)*(x^2)*exp(2*x)+(-1737019/92823250)*cos(3*x)+(-4850123/556139500)*sin(3*x)+ yp(x)
# La solución general es:
show(y)
x ↦ 10202197 x2e(2 x)−110 x2ex−11028196 (3 (598 x+753)cos⁡(3 x)+(351 x−2758)sin⁡(3 x))x−4333028561 xe(2 x)−1425 xex−173701992823250 cos⁡(3 x)+774782371293 e(2 x)−267250 ex−4850123556139500 sin⁡(3 x)\displaystyle x \ {\mapsto}\ \frac{1020}{2197} \, x^{2} e^{\left(2 \, x\right)} - \frac{1}{10} \, x^{2} e^{x} - \frac{1}{1028196} \, {\left(3 \, {\left(598 \, x + 753\right)} \cos\left(3 \, x\right) + {\left(351 \, x - 2758\right)} \sin\left(3 \, x\right)\right)} x - \frac{43330}{28561} \, x e^{\left(2 \, x\right)} - \frac{14}{25} \, x e^{x} - \frac{1737019}{92823250} \, \cos\left(3 \, x\right) + \frac{774782}{371293} \, e^{\left(2 \, x\right)} - \frac{267}{250} \, e^{x} - \frac{4850123}{556139500} \, \sin\left(3 \, x\right)

problema 3 encontrar los primeros seis terminos de cada una de las soluciones linealmente independientes de la forma ΣCn Xn



#Generando una sumatoria del 12 terminos 
reset()
n=12
b= list (var('a%d' % i) for i in range (n) )
x= var('x')
y= function('y')(x)
var('a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11')

a0=0
a1=1
y(x)=a0+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11
show(y(x))
a11x11+a10x10+a9x9+a8x8+a7x7+a6x6+a5x5+a4x4+a3x3+a2x2+x\displaystyle a_{11} x^{11} + a_{10} x^{10} + a_{9} x^{9} + a_{8} x^{8} + a_{7} x^{7} + a_{6} x^{6} + a_{5} x^{5} + a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + x
#definir la serie de seno para poder introducirla en la ecuacion diferencial.
senoX=taylor(sin(x),x,0,12) #para solo mostar los primeros terminos de la serie del seno
show(senoX)
#sustituyendo en la ecuancion diferencial y expandiendo tenemos
a0=0
a1=1
ecua=expand(x*diff(y,x,2)+senoX*diff(y,x,1)+x*y(x)==0)
−139916800 x11+1362880 x9−15040 x7+1120 x5−16 x3+x\displaystyle -\frac{1}{39916800} \, x^{11} + \frac{1}{362880} \, x^{9} - \frac{1}{5040} \, x^{7} + \frac{1}{120} \, x^{5} - \frac{1}{6} \, x^{3} + x

show(ecua)
x ↦ −13628800 a11x21−13991680 a10x20+11362880 a11x19−14435200 a9x19+136288 a10x18−14989600 a8x18−115040 a11x17−15702400 a7x17+140320 a9x17−1504 a10x16−16652800 a6x16+145360 a8x16+11120 a11x15−17983360 a5x15+151840 a7x15−1560 a9x15+112 a10x14−19979200 a4x14+160480 a6x14−1630 a8x14−116 a11x13−113305600 a3x13+172576 a5x13−1720 a7x13+340 a9x13−53 a10x12+a11x12−119958400 a2x12+190720 a4x12−1840 a6x12+115 a8x12+a10x11+11 a11x11+1120960 a3x11−11008 a5x11+7120 a7x11−32 a9x11+10 a10x10+110 a11x10+1181440 a2x10−11260 a4x10+120 a6x10−43 a8x10+a9x10−139916800 x11+90 a10x9−11680 a3x9+124 a5x9−76 a7x9+a8x9+9 a9x9−12520 a2x8+130 a4x8−a6x8+a7x8+8 a8x8+72 a9x8+1362880 x9+140 a3x7−56 a5x7+a6x7+7 a7x7+56 a8x7+160 a2x6−23 a4x6+a5x6+6 a6x6+42 a7x6−15040 x7−12 a3x5+a4x5+5 a5x5+30 a6x5−13 a2x4+a3x4+4 a4x4+20 a5x4+1120 x5+a2x3+3 a3x3+12 a4x3+2 a2x2+6 a3x2−16 x3+2 a2x+x2+x=0\displaystyle x \ {\mapsto}\ -\frac{1}{3628800} \, a_{11} x^{21} - \frac{1}{3991680} \, a_{10} x^{20} + \frac{11}{362880} \, a_{11} x^{19} - \frac{1}{4435200} \, a_{9} x^{19} + \frac{1}{36288} \, a_{10} x^{18} - \frac{1}{4989600} \, a_{8} x^{18} - \frac{11}{5040} \, a_{11} x^{17} - \frac{1}{5702400} \, a_{7} x^{17} + \frac{1}{40320} \, a_{9} x^{17} - \frac{1}{504} \, a_{10} x^{16} - \frac{1}{6652800} \, a_{6} x^{16} + \frac{1}{45360} \, a_{8} x^{16} + \frac{11}{120} \, a_{11} x^{15} - \frac{1}{7983360} \, a_{5} x^{15} + \frac{1}{51840} \, a_{7} x^{15} - \frac{1}{560} \, a_{9} x^{15} + \frac{1}{12} \, a_{10} x^{14} - \frac{1}{9979200} \, a_{4} x^{14} + \frac{1}{60480} \, a_{6} x^{14} - \frac{1}{630} \, a_{8} x^{14} - \frac{11}{6} \, a_{11} x^{13} - \frac{1}{13305600} \, a_{3} x^{13} + \frac{1}{72576} \, a_{5} x^{13} - \frac{1}{720} \, a_{7} x^{13} + \frac{3}{40} \, a_{9} x^{13} - \frac{5}{3} \, a_{10} x^{12} + a_{11} x^{12} - \frac{1}{19958400} \, a_{2} x^{12} + \frac{1}{90720} \, a_{4} x^{12} - \frac{1}{840} \, a_{6} x^{12} + \frac{1}{15} \, a_{8} x^{12} + a_{10} x^{11} + 11 \, a_{11} x^{11} + \frac{1}{120960} \, a_{3} x^{11} - \frac{1}{1008} \, a_{5} x^{11} + \frac{7}{120} \, a_{7} x^{11} - \frac{3}{2} \, a_{9} x^{11} + 10 \, a_{10} x^{10} + 110 \, a_{11} x^{10} + \frac{1}{181440} \, a_{2} x^{10} - \frac{1}{1260} \, a_{4} x^{10} + \frac{1}{20} \, a_{6} x^{10} - \frac{4}{3} \, a_{8} x^{10} + a_{9} x^{10} - \frac{1}{39916800} \, x^{11} + 90 \, a_{10} x^{9} - \frac{1}{1680} \, a_{3} x^{9} + \frac{1}{24} \, a_{5} x^{9} - \frac{7}{6} \, a_{7} x^{9} + a_{8} x^{9} + 9 \, a_{9} x^{9} - \frac{1}{2520} \, a_{2} x^{8} + \frac{1}{30} \, a_{4} x^{8} - a_{6} x^{8} + a_{7} x^{8} + 8 \, a_{8} x^{8} + 72 \, a_{9} x^{8} + \frac{1}{362880} \, x^{9} + \frac{1}{40} \, a_{3} x^{7} - \frac{5}{6} \, a_{5} x^{7} + a_{6} x^{7} + 7 \, a_{7} x^{7} + 56 \, a_{8} x^{7} + \frac{1}{60} \, a_{2} x^{6} - \frac{2}{3} \, a_{4} x^{6} + a_{5} x^{6} + 6 \, a_{6} x^{6} + 42 \, a_{7} x^{6} - \frac{1}{5040} \, x^{7} - \frac{1}{2} \, a_{3} x^{5} + a_{4} x^{5} + 5 \, a_{5} x^{5} + 30 \, a_{6} x^{5} - \frac{1}{3} \, a_{2} x^{4} + a_{3} x^{4} + 4 \, a_{4} x^{4} + 20 \, a_{5} x^{4} + \frac{1}{120} \, x^{5} + a_{2} x^{3} + 3 \, a_{3} x^{3} + 12 \, a_{4} x^{3} + 2 \, a_{2} x^{2} + 6 \, a_{3} x^{2} - \frac{1}{6} \, x^{3} + 2 \, a_{2} x + x^{2} + x = 0
#agrupando los coeficientes de x^1 y resolviendo el sistema para encontra a2
solve([(ecua.lhs()).coefficient(x,1)==0],a2)
show(solve([(ecua.lhs()).coefficient(x,1)==0],a2))
[a2 == (-1/2)]
[a2=(−12)\displaystyle a_{2} = \left(-\frac{1}{2}\right)]
#agrupando coeficientes de x^2 y resolviendo para encontrar a3
show(solve([(ecua.lhs()).coefficient(x,2).substitute(a2=-1/2)==0],a3))
[a3=0\displaystyle a_{3} = 0]

#agrupando coeficientes de x^3 y resolviendo para encontrar a4
show(solve([(ecua.lhs()).coefficient(x,3).substitute(a2=-1/2,a3=0)==0],a4))
[a4=(118)\displaystyle a_{4} = \left(\frac{1}{18}\right)]

#agrupando coeficientes de x^4 y resolviendo para encontrar a5
show(solve([(ecua.lhs()).coefficient(x,4).substitute(a2=-1/2,a3=0,a4=1/18)==0],a5))
[a5=(−7360)\displaystyle a_{5} = \left(-\frac{7}{360}\right)]

#agrupando coeficientes de x^5 y resolviendo para encontrar a6
show(solve([(ecua.lhs()).coefficient(x,5).substitute(a2=-1/2,a3=0,a4=1/18,a5=-7/360)==0],a6))
[a6=(1900)\displaystyle a_{6} = \left(\frac{1}{900}\right)]

#agrupando coeficientes de x^6 y resolviendo para encontrar a7
show(solve([(ecua.lhs()).coefficient(x,6).substitute(a2=-1/2,a3=0,a4=1/18,a5=-7/360,a6=1/900)==0],a7))
[a7=(157113400)\displaystyle a_{7} = \left(\frac{157}{113400}\right)]

#agrupando coeficientes de x^7 y resolviendo para encontrar a8
show(solve([(ecua.lhs()).coefficient(x,7).substitute(a2=-1/2,a3=0,a4=1/18,a5=-7/360,a6=1/900,a7=157/113400)==0],a8))
[a8=(−1939690)\displaystyle a_{8} = \left(-\frac{19}{39690}\right)]

#agrupando coeficientes de x^8 y resolviendo para encontrar a9
show(solve([(ecua.lhs()).coefficient(x,8).substitute(a2=-1/2,a3=0,a4=1/18,a5=-7/360,a6=1/900,a7=157/113400,a8=-19/39690)==0],a9))
[a9=(79738102400)\displaystyle a_{9} = \left(\frac{797}{38102400}\right)]

#agrupando coeficientes de x^9 y resolviendo para encontrar a10
show(solve([(ecua.lhs()).coefficient(x,9).substitute(a2=-1/2,a3=0,a4=1/18,a5=-7/360,a6=1/900,a7=157/113400,a8=-19/39690,a9=797/38102400)==0],a10))
[a10=(92330618000)\displaystyle a_{10} = \left(\frac{923}{30618000}\right)]
#agrupando coeficientes de x^10 y resolviendo para encontrar a11
show(solve([(ecua.lhs()).coefficient(x,10).substitute(a2=-1/2,a3=0,a4=1/18,a5=-7/360,a6=1/900,a7=157/113400,a8=-19/39690,a9=797/38102400,a10=923/30618000)==0],a11))
[a11=(−41551947151720000)\displaystyle a_{11} = \left(-\frac{415519}{47151720000}\right)]
# ahora que ya caculamos los ao, c1,a2,... a11 los sustituimos y tenemos que nuestra primera solucion independiente es y1(x), esto se tiene debido a que y=0 es punto ordinario y debido a un teorema visto en clase que nos asegura que habran dos soluciones linealmente independiente
y1(x)=0+1*x+(-1/2)*x^2+(0)*x^3+(1/18)*x^4+(-7/360)*x^5+(1/900)*x^6+(157/113400)*x^7+(-1939690)*x^8+(797/38102400)*x^9+(923/30618000)*x^10+(-415519/47151720000)*x^11
show(y1(x))
−41551947151720000 x11+92330618000 x10+79738102400 x9−1939690 x8+157113400 x7+1900 x6−7360 x5+118 x4−12 x2+x\displaystyle -\frac{415519}{47151720000} \, x^{11} + \frac{923}{30618000} \, x^{10} + \frac{797}{38102400} \, x^{9} - 1939690 \, x^{8} + \frac{157}{113400} \, x^{7} + \frac{1}{900} \, x^{6} - \frac{7}{360} \, x^{5} + \frac{1}{18} \, x^{4} - \frac{1}{2} \, x^{2} + x

#para la segunda solucion  tenemos
#Generando una sumatoria del 12 terminos 

n=12
b= list (var('a%d' % i) for i in range (n) )
x= var('x')
y= function('y')(x)

var('a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11')
a0=1
a1=0
y(x)=a0+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11
show(y(x))
#definir la serie de seno para poder introducirla en la ecuacion diferencial.
senoX=taylor(sin(x),x,0,12) #para solo mostar los primeros terminos de la serie del seno
show(senoX)
#sustituyendo en la ecuancion diferencial y expandiendo tenemos
a0=1
a1=0
ecua=expand(x*diff(y,x,2)+senoX*diff(y,x,1)+x*y(x)==0)
show(ecua)
(a0, a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, a11)
a11x11+a10x10+a9x9+a8x8+a7x7+a6x6+a5x5+a4x4+a3x3+a2x2+1\displaystyle a_{11} x^{11} + a_{10} x^{10} + a_{9} x^{9} + a_{8} x^{8} + a_{7} x^{7} + a_{6} x^{6} + a_{5} x^{5} + a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + 1
−139916800 x11+1362880 x9−15040 x7+1120 x5−16 x3+x\displaystyle -\frac{1}{39916800} \, x^{11} + \frac{1}{362880} \, x^{9} - \frac{1}{5040} \, x^{7} + \frac{1}{120} \, x^{5} - \frac{1}{6} \, x^{3} + x
x ↦ −13628800 a11x21−13991680 a10x20+11362880 a11x19−14435200 a9x19+136288 a10x18−14989600 a8x18−115040 a11x17−15702400 a7x17+140320 a9x17−1504 a10x16−16652800 a6x16+145360 a8x16+11120 a11x15−17983360 a5x15+151840 a7x15−1560 a9x15+112 a10x14−19979200 a4x14+160480 a6x14−1630 a8x14−116 a11x13−113305600 a3x13+172576 a5x13−1720 a7x13+340 a9x13−53 a10x12+a11x12−119958400 a2x12+190720 a4x12−1840 a6x12+115 a8x12+a10x11+11 a11x11+1120960 a3x11−11008 a5x11+7120 a7x11−32 a9x11+10 a10x10+110 a11x10+1181440 a2x10−11260 a4x10+120 a6x10−43 a8x10+a9x10+90 a10x9−11680 a3x9+124 a5x9−76 a7x9+a8x9+9 a9x9−12520 a2x8+130 a4x8−a6x8+a7x8+8 a8x8+72 a9x8+140 a3x7−56 a5x7+a6x7+7 a7x7+56 a8x7+160 a2x6−23 a4x6+a5x6+6 a6x6+42 a7x6−12 a3x5+a4x5+5 a5x5+30 a6x5−13 a2x4+a3x4+4 a4x4+20 a5x4+a2x3+3 a3x3+12 a4x3+2 a2x2+6 a3x2+2 a2x+x=0\displaystyle x \ {\mapsto}\ -\frac{1}{3628800} \, a_{11} x^{21} - \frac{1}{3991680} \, a_{10} x^{20} + \frac{11}{362880} \, a_{11} x^{19} - \frac{1}{4435200} \, a_{9} x^{19} + \frac{1}{36288} \, a_{10} x^{18} - \frac{1}{4989600} \, a_{8} x^{18} - \frac{11}{5040} \, a_{11} x^{17} - \frac{1}{5702400} \, a_{7} x^{17} + \frac{1}{40320} \, a_{9} x^{17} - \frac{1}{504} \, a_{10} x^{16} - \frac{1}{6652800} \, a_{6} x^{16} + \frac{1}{45360} \, a_{8} x^{16} + \frac{11}{120} \, a_{11} x^{15} - \frac{1}{7983360} \, a_{5} x^{15} + \frac{1}{51840} \, a_{7} x^{15} - \frac{1}{560} \, a_{9} x^{15} + \frac{1}{12} \, a_{10} x^{14} - \frac{1}{9979200} \, a_{4} x^{14} + \frac{1}{60480} \, a_{6} x^{14} - \frac{1}{630} \, a_{8} x^{14} - \frac{11}{6} \, a_{11} x^{13} - \frac{1}{13305600} \, a_{3} x^{13} + \frac{1}{72576} \, a_{5} x^{13} - \frac{1}{720} \, a_{7} x^{13} + \frac{3}{40} \, a_{9} x^{13} - \frac{5}{3} \, a_{10} x^{12} + a_{11} x^{12} - \frac{1}{19958400} \, a_{2} x^{12} + \frac{1}{90720} \, a_{4} x^{12} - \frac{1}{840} \, a_{6} x^{12} + \frac{1}{15} \, a_{8} x^{12} + a_{10} x^{11} + 11 \, a_{11} x^{11} + \frac{1}{120960} \, a_{3} x^{11} - \frac{1}{1008} \, a_{5} x^{11} + \frac{7}{120} \, a_{7} x^{11} - \frac{3}{2} \, a_{9} x^{11} + 10 \, a_{10} x^{10} + 110 \, a_{11} x^{10} + \frac{1}{181440} \, a_{2} x^{10} - \frac{1}{1260} \, a_{4} x^{10} + \frac{1}{20} \, a_{6} x^{10} - \frac{4}{3} \, a_{8} x^{10} + a_{9} x^{10} + 90 \, a_{10} x^{9} - \frac{1}{1680} \, a_{3} x^{9} + \frac{1}{24} \, a_{5} x^{9} - \frac{7}{6} \, a_{7} x^{9} + a_{8} x^{9} + 9 \, a_{9} x^{9} - \frac{1}{2520} \, a_{2} x^{8} + \frac{1}{30} \, a_{4} x^{8} - a_{6} x^{8} + a_{7} x^{8} + 8 \, a_{8} x^{8} + 72 \, a_{9} x^{8} + \frac{1}{40} \, a_{3} x^{7} - \frac{5}{6} \, a_{5} x^{7} + a_{6} x^{7} + 7 \, a_{7} x^{7} + 56 \, a_{8} x^{7} + \frac{1}{60} \, a_{2} x^{6} - \frac{2}{3} \, a_{4} x^{6} + a_{5} x^{6} + 6 \, a_{6} x^{6} + 42 \, a_{7} x^{6} - \frac{1}{2} \, a_{3} x^{5} + a_{4} x^{5} + 5 \, a_{5} x^{5} + 30 \, a_{6} x^{5} - \frac{1}{3} \, a_{2} x^{4} + a_{3} x^{4} + 4 \, a_{4} x^{4} + 20 \, a_{5} x^{4} + a_{2} x^{3} + 3 \, a_{3} x^{3} + 12 \, a_{4} x^{3} + 2 \, a_{2} x^{2} + 6 \, a_{3} x^{2} + 2 \, a_{2} x + x = 0

#agrupando los coeficientes de x^1 y resolviendo el sistema para encontra a2
solve([(ecua.lhs()).coefficient(x,1)==0],a2)
show(solve([(ecua.lhs()).coefficient(x,1)==0],a2))
[a2 == (-1/2)]
[a2=(−12)\displaystyle a_{2} = \left(-\frac{1}{2}\right)]
#agrupando coeficientes de x^2 y resolviendo para encontrar a3
show(solve([(ecua.lhs()).coefficient(x,2).substitute(a2=-1/2)==0],a3))
[a3=(16)\displaystyle a_{3} = \left(\frac{1}{6}\right)]
#agrupando coeficientes de x^3 y resolviendo para encontrar a4
show(solve([(ecua.lhs()).coefficient(x,3).substitute(a2=-1/2,a3=1/6)==0],a4))
[a4=0\displaystyle a_{4} = 0]
#agrupando coeficientes de x^4 y resolviendo para encontrar a5
show(solve([(ecua.lhs()).coefficient(x,4).substitute(a2=-1/2,a3=1/6,a4=0)==0],a5))
[a5=(−160)\displaystyle a_{5} = \left(-\frac{1}{60}\right)]

#agrupando coeficientes de x^5 y resolviendo para encontrar a6
show(solve([(ecua.lhs()).coefficient(x,5).substitute(a2=-1/2,a3=1/6,a4=0,a5=-1/60)==0],a6))
[a6=(1180)\displaystyle a_{6} = \left(\frac{1}{180}\right)]
#agrupando coeficientes de x^6 y resolviendo para encontrar a7
show(solve([(ecua.lhs()).coefficient(x,6).substitute(a2=-1/2,a3=1/6,a4=0,a5=-1/60,a6=1/180)==0],a7))
[a7=(−15040)\displaystyle a_{7} = \left(-\frac{1}{5040}\right)]
#agrupando coeficientes de x^7 y resolviendo para encontrar a8
show(solve([(ecua.lhs()).coefficient(x,7).substitute(a2=-1/2,a3=1/6,a4=0,a5=-1/60,a6=1/180,a7=-1/5040)==0],a8))
[a8=(−12520)\displaystyle a_{8} = \left(-\frac{1}{2520}\right)]
#agrupando coeficientes de x^8y resolviendo para encontrar a9
show(solve([(ecua.lhs()).coefficient(x,8).substitute(a2=-1/2,a3=1/6,a4=0,a5=-1/60,a6=1/180,a7=-1/5040,a8=-1/2520)==0],a9))
[a9=(1190720)\displaystyle a_{9} = \left(\frac{11}{90720}\right)]

#agrupando coeficientes de x^9y resolviendo para encontrar a10
show(solve([(ecua.lhs()).coefficient(x,9).substitute(a2=-1/2,a3=1/6,a4=0,a5=-1/60,a6=1/180,a7=-1/5040,a8=-1/2520,a9=11/90720)==0],a10))
[a10=(−1680400)\displaystyle a_{10} = \left(-\frac{1}{680400}\right)]

#agrupando coeficientes de x^10 y resolviendo para encontrar a11
show(solve([(ecua.lhs()).coefficient(x,10).substitute(a2=-1/2,a3=1/6,a4=0,a5=-1/60,a6=1/180,a7=-1/5040,a8=-1/2520,a9=11/90720,a10=-1/680400)==0],a11))
[a11=(−4957598752000)\displaystyle a_{11} = \left(-\frac{4957}{598752000}\right)]

#ahora que ya calculamos los nuevos valores para a0,a1,a2,....a11, los sustituimos para obtener nuestra segunda solucion
y2(x)=1+0*x+(-1/2)*x^2+(1/6)*x^3+(0)*x^4+(-1/60)*x^5+(1/180)*x^6+(-1/5040)*x^7+(-1/2520)*x^8+(11/90720)*x^9+(-1/680400)*x^10+(-4957/598752000)*x^11
show(y2(x))
−4957598752000 x11−1680400 x10+1190720 x9−12520 x8−15040 x7+1180 x6−160 x5+16 x3−12 x2+1\displaystyle -\frac{4957}{598752000} \, x^{11} - \frac{1}{680400} \, x^{10} + \frac{11}{90720} \, x^{9} - \frac{1}{2520} \, x^{8} - \frac{1}{5040} \, x^{7} + \frac{1}{180} \, x^{6} - \frac{1}{60} \, x^{5} + \frac{1}{6} \, x^{3} - \frac{1}{2} \, x^{2} + 1