Integration of rational function by partial functions. The last step confuses me!

  • Thread starter randoreds
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  • #1
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Ok all save y'all a little reading.
Worked out the problem. Got
X^2+2x-1=A(2x-1)(x+2)+bx(x+2)+cx(2x-1)

Ok then you write it standard form for a polynomial. Then use can use there coefficients to write new equations at you get

2a+b+c=1
3a+2b-c=2 and finally
-2a=1

Now you solve for a,b,c and this is where I'm confused.
To solve for a is easy, it is just 1/2. But the book doesn't show how to solve for b and c.
B =1/5 and C =-1/10

I'm totally confused how they got those numbers so if you could explain it. That would be so helpful!!
 

Answers and Replies

  • #2
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This is a 3x3 system of linear equations. Are you saying you don't know how to solve it?
 
  • #3
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This is a 3x3 system of linear equations. Are you saying you don't know how to solve it?
I guess yes. I totally don't remember how to solve these
 
  • #4
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But thanks, Ill look it up. I couldn't remember what they were called. Thanks
 
  • #6
HallsofIvy
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Ok all save y'all a little reading.
Worked out the problem. Got
X^2+2x-1=A(2x-1)(x+2)+bx(x+2)+cx(2x-1)

Ok then you write it standard form for a polynomial. Then use can use there coefficients to write new equations at you get

2a+b+c=1
3a+2b-c=2 and finally
-2a=1
This last is incorrect. It should be -2a= -1 or 2a= 1. That's why a= 1/2 rather than -1/2.

Now you solve for a,b,c and this is where I'm confused.
To solve for a is easy, it is just 1/2. But the book doesn't show how to solve for b and c.
B =1/5 and C =-1/10
Putting a= 1/2 into the other two equations, 2a+ b+ c= 1+ b+ c= 1 so b+ c= 0 and 3a+ 2b- c= 3/2+ 2b- c= 2 so that 2b- c= 1/2. If you don't like fractions multiply both sides by 2 to get 4b- c= 1. From b+ c= 0, b= -c so 4b= -4c. 4b- c= -4c- c= -5c= 7.

Of course, you can set each corresponding coefficients equal because x^2+2x-1=a(2x-1)(x+2)+bx(x+2)+cx(2x-1) is true for all x. So you can get the three equations to solve for a, b, and c by taking three values for x. And choosing those value cleverly can simplify the resulting equations! For example, if you take x= 0, x(x+2) and x(2x-1) are both equal to 0. The equation, for x= 0, becomes 0^2+ 2(0)-1= a(2(0)-1)(0+2)+ b(0)+ c(0)= -2a. That is, -2a= -1 or 2a= 1. Taking x= -2, both (2x-1)(x+2) and x(x+2) are 0 so the equation becomes 2^2- 2(2)- 1= a(0)+ b(0)+c(2)(2(2)-1) or 6c= -1. Finally, if you take x= 1/2, (2x-1)(x+2) and x(2x-1) are 0 and the equation is (1/2)^2+2(1/2)- 1= b(1/2)(1/2+ 2) so 1/4= (5/4)b.

I'm totally confused how they got those numbers so if you could explain it. That would be so helpful!!
 

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