- #1
phy$x
- 3
- 0
I've been trying to get out this question for a while now:
ai) Show that (x,y,z) = (1,1,1) is a solution to the following system of equations:
x + y + z = 3
2x + 2y + 2z = 6
3x + 3y +3z = 9
aii) Hence find the general solution of the system
b) Express 2x^2 + 3/(x^2 + 1)^2 in partial fractions
My attempt:
Well ai) was simple and i got that part out with barely any effort.
In aii), i don't even know how to start :( All i know is that the answer is supposed to be:
(x,y,x) = λ(1,0,-1) + μ(0,1,-1) + (1,1,1,)
Sorry i can't offer any attempt...i just really don't know where to start...any help at all will be appreciated here.
With b) i used the matrix method...but that wasnt the approach they were looking for. I was supposed to use the concept of repeated factors:
2x^2 + 3/(x^2 + 1)^2 = Ax + B/x^2 + 1 + CX + D/(x^2+1)^2
(multiply throughout by (x^2 + 1)^2)
2x^2 +3 = (Ax + B)(x^2 + 1) + Cx + D
Let x=0
3 = B + D
D = 3 - B
Let x= 1
5 = (A + B)(2) + C + D
5 = 2A + 2B + C + D
Substituting D = 3 - B
5 = 2A + 2B + C + 3 - B
2 = 2A + B + C
Well this is where I am stuck...any help at all would be a life saver. Thank you in advance.
ai) Show that (x,y,z) = (1,1,1) is a solution to the following system of equations:
x + y + z = 3
2x + 2y + 2z = 6
3x + 3y +3z = 9
aii) Hence find the general solution of the system
b) Express 2x^2 + 3/(x^2 + 1)^2 in partial fractions
My attempt:
Well ai) was simple and i got that part out with barely any effort.
In aii), i don't even know how to start :( All i know is that the answer is supposed to be:
(x,y,x) = λ(1,0,-1) + μ(0,1,-1) + (1,1,1,)
Sorry i can't offer any attempt...i just really don't know where to start...any help at all will be appreciated here.
With b) i used the matrix method...but that wasnt the approach they were looking for. I was supposed to use the concept of repeated factors:
2x^2 + 3/(x^2 + 1)^2 = Ax + B/x^2 + 1 + CX + D/(x^2+1)^2
(multiply throughout by (x^2 + 1)^2)
2x^2 +3 = (Ax + B)(x^2 + 1) + Cx + D
Let x=0
3 = B + D
D = 3 - B
Let x= 1
5 = (A + B)(2) + C + D
5 = 2A + 2B + C + D
Substituting D = 3 - B
5 = 2A + 2B + C + 3 - B
2 = 2A + B + C
Well this is where I am stuck...any help at all would be a life saver. Thank you in advance.