Proyecto Mate 4 P1 (Saraí Perez).sagews

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## Problem 1
### Solve the following diferential equation subject to the initial conditions given:
$(D-2)^3 (D^2 +9)y=x^2 e^x + xsin (3x) \  ; \  y(0)=y'(0)=y"(0)=y^(3)*(0)=y^(4)*(0)=1$

Problem 1

Solve the following diferential equation subject to the initial conditions given:

$(D-2)^3 (D^2 +9)y=x^2 e^x + xsin (3x) \ ; \ y(0)=y'(0)=y"(0)=y^(3)*(0)=y^(4)*(0)=1$

# Let us first find the characteristic equation and solve it:
var('r')
solve([(r-2)^3*(r^2 + 9)== 0],r)

r [r == (-3*I), r == (3*I), r == 2]

# Among the roots from the characteristic equations that were found, there is one with multiplicity 3 and this one will be used to find the complementary solution. Let's start with the particular one:
#We have multiplied everything times x to take out the duplicity.
var ('x,A,B,C,D,E,F,G')
yp=(A+B*x+C*x^2)*exp(x)+(D*x+E*x^2)*cos(3*x)+(F*x+G*x^2)*sin(3*x)
show(yp)

(x, A, B, C, D, E, F, G)

$\displaystyle {\left(E x^{2} + D x\right)} \cos\left(3 \, x\right) + {\left(C x^{2} + B x + A\right)} e^{x} + {\left(G x^{2} + F x\right)} \sin\left(3 \, x\right)$

#Now, we will substitute the previous equation into the original differential equation:
rem0= diff(yp,x,2)+9*yp
rem1= diff(rem0,x)-2*rem0
rem2= diff(rem1,x)-2*rem1
rem3= diff(rem2,x)-2*rem2
remfinal= rem3-x^2*exp(x)-x*sin(3*x)
show(remfinal)

$\displaystyle -x^{2} e^{x} - 54 \, {\left(2 \, E x + D\right)} \cos\left(3 \, x\right) + 276 \, {\left(2 \, G x + F\right)} \cos\left(3 \, x\right) + 524 \, E \cos\left(3 \, x\right) - 162 \, G \cos\left(3 \, x\right) - 10 \, {\left(C x^{2} + B x + A\right)} e^{x} + 28 \, {\left(2 \, C x + B\right)} e^{x} - 50 \, C e^{x} - 276 \, {\left(2 \, E x + D\right)} \sin\left(3 \, x\right) - 54 \, {\left(2 \, G x + F\right)} \sin\left(3 \, x\right) + 162 \, E \sin\left(3 \, x\right) + 524 \, G \sin\left(3 \, x\right) - x \sin\left(3 \, x\right)$

#Now, we will find the coefficients of the previous equation
coeffEx=remfinal.coefficient(exp(x))
coeffCos=remfinal.coefficient(cos(3*x))
coeffSin=remfinal.coefficient(sin(3*x))

#We will use the command Solve to find the coefficients A, B, C.
solve([coeffEx.coefficient(x,0)==0,coeffEx.coefficient(x,1)==0,
coeffEx.coefficient(x,2)==0],A,B,C)

[[A == (-267/250), B == (-14/25), C == (-1/10)]]

#And now, we will find the remaining coefficients of the problem, which are the coefficients D, E, F, G.
solve ([coeffCos.coefficient(x,0)==0,coeffCos.coefficient(x,1)==0,
coeffSin.coefficient(x,0)==0,coeffSin.coefficient(x,1)==0],D,E,F,G)

[[D == (-251/114244), E == (-23/13182), F == (1379/514098), G == (-3/8788)]]

#In short, the values of the coefficients found are the following ones:
A=-267/250; B=-14/25; C=-1/10; D=-251/114244; E=-23/13182; F=1379/514098; G=-3/8788;
#We are now finally able to find the complementary solution as follows:
var ('c1,c2,c3,c4,c5')
yc=c1*exp(2*x)+c2 *x*exp(2*x)+c3*x^2*exp(2*x)+c4*cos(3*x)+c5*sin(3*x)+(A+B*x+C*x^2)*exp(x)+(D*x+E*x^2)*cos(3*x)+(F*x+G*x^2)*sin(3*x)
show(yc)

(c1, c2, c3, c4, c5)

$\displaystyle c_{3} x^{2} e^{\left(2 \, x\right)} + c_{2} x e^{\left(2 \, x\right)} - \frac{1}{342732} \, {\left(598 \, x^{2} + 753 \, x\right)} \cos\left(3 \, x\right) + c_{4} \cos\left(3 \, x\right) + c_{1} e^{\left(2 \, x\right)} - \frac{1}{250} \, {\left(25 \, x^{2} + 140 \, x + 267\right)} e^{x} - \frac{1}{1028196} \, {\left(351 \, x^{2} - 2758 \, x\right)} \sin\left(3 \, x\right) + c_{5} \sin\left(3 \, x\right)$

#Solving for the coeffients c1,  c2, c3, c4, c5 we find:
solve([yc.substitute(x=0)==1,diff(yc,x).substitute(x=0)==1,diff(yc,x,2).substitute(x=0)==1,diff(yc,x,3).substitute(x=0)==1,diff(yc,x,4).substitute(x=0)==1],c1,c2,c3,c4,c5)

[[c1 == (774782/371293), c2 == (-43330/28561), c3 == (1020/2197), c4 == (-1737019/92823250), c5 == (-4850123/556939500)]]

#In short, the values of the coefficients c1, c2, c3, c4, c5 are:
c1 =774782/371293; c2 = -43330/28561; c3 =1020/2197; c4= -1737019/92823250; c5 = -4850123/556939500
#And my complete solution (complementary and particular)for the differential equation with the initial conditions given, is:
yc=c1*exp(2*x)+c2*x*exp(2*x)+c3*x^2*exp(2*x)+c4*cos(3*x)+c5*sin(3*x)+(A+B*x+C*x^2)*exp(x)+(D*x+E*x^2)*cos(3*x)+(F*x+G*x^2)*sin(3*x)
show ( yc )

$\displaystyle \frac{1020}{2197} \, x^{2} e^{\left(2 \, x\right)} - \frac{1}{342732} \, {\left(598 \, x^{2} + 753 \, x\right)} \cos\left(3 \, x\right) - \frac{43330}{28561} \, x e^{\left(2 \, x\right)} - \frac{1}{250} \, {\left(25 \, x^{2} + 140 \, x + 267\right)} e^{x} - \frac{1}{1028196} \, {\left(351 \, x^{2} - 2758 \, x\right)} \sin\left(3 \, x\right) - \frac{1737019}{92823250} \, \cos\left(3 \, x\right) + \frac{774782}{371293} \, e^{\left(2 \, x\right)} - \frac{4850123}{556939500} \, \sin\left(3 \, x\right)$