`x^3/((x+2)^2(x-2)^2)`
Let `x^3/((x+2)^2(x-2)^2)=A/(x+2)+B/(x+2)^2+C/(x-2)+D/(x-2)^2`
`x^3/((x+2)^2(x-2)^2)=(A(x+2)(x-2)^2+B(x-2)^2+C(x-2)(x+2)^2+D(x+2)^2)/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(A(x+2)(x^2-4x+4)+B(x^2-4x+4)+C(x-2)(x^2+4x+4)+D(x^2+4x+4))/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(A(x^3-4x^2+4x+2x^2-8x+8)+B(x^2-4x+4)+C(x^3+4x^2+4x-2x^2-8x-8)+D(x^2+4x+4))/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(A(x^3-2x^2-4x+8)+B(x^2-4x+4)+C(x^3+2x^2-4x-8)+D(x^2+4x+4))/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(x^3(A+C)+x^2(-2A+B+2C+D)+x(-4A-4B-4C+4D)+8A+4B-8C+4D)/((x+2)^2(x-2)^2)`
`:.x^3=x^3(A+C)+x^2(-2A+B+2C+D)+x(-4A-4B-4C+4D)+8A+4B-8C+4D`
equating the coefficients of the like terms,
`A+C=1`
`-2A+B+2C+D=0`
`-4A-4B-4C+4D=0`
`8A+4B-8C+4D=0`
Now we have to solve the above four equations to get the solutions of A,B,C and D.
Rewrite the third equation as,
`-4(A+C)-4B+4D=0`
substitute the expression of ( A+C) from the first equation in the above equation,
`-4(1)-4B+4D=0`
`-4B+4D=4`
`4(-B+D)=4`
`-B+D=1`
`D=1+B`
Express C in terms of A from first equation,
`C=1-A`
Substitute the expressions...
`x^3/((x+2)^2(x-2)^2)`
Let `x^3/((x+2)^2(x-2)^2)=A/(x+2)+B/(x+2)^2+C/(x-2)+D/(x-2)^2`
`x^3/((x+2)^2(x-2)^2)=(A(x+2)(x-2)^2+B(x-2)^2+C(x-2)(x+2)^2+D(x+2)^2)/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(A(x+2)(x^2-4x+4)+B(x^2-4x+4)+C(x-2)(x^2+4x+4)+D(x^2+4x+4))/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(A(x^3-4x^2+4x+2x^2-8x+8)+B(x^2-4x+4)+C(x^3+4x^2+4x-2x^2-8x-8)+D(x^2+4x+4))/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(A(x^3-2x^2-4x+8)+B(x^2-4x+4)+C(x^3+2x^2-4x-8)+D(x^2+4x+4))/((x+2)^2(x-2)^2)`
`x^3/((x+2)^2(x-2)^2)=(x^3(A+C)+x^2(-2A+B+2C+D)+x(-4A-4B-4C+4D)+8A+4B-8C+4D)/((x+2)^2(x-2)^2)`
`:.x^3=x^3(A+C)+x^2(-2A+B+2C+D)+x(-4A-4B-4C+4D)+8A+4B-8C+4D`
equating the coefficients of the like terms,
`A+C=1`
`-2A+B+2C+D=0`
`-4A-4B-4C+4D=0`
`8A+4B-8C+4D=0`
Now we have to solve the above four equations to get the solutions of A,B,C and D.
Rewrite the third equation as,
`-4(A+C)-4B+4D=0`
substitute the expression of ( A+C) from the first equation in the above equation,
`-4(1)-4B+4D=0`
`-4B+4D=4`
`4(-B+D)=4`
`-B+D=1`
`D=1+B`
Express C in terms of A from first equation,
`C=1-A`
Substitute the expressions of C and D in the second equation,
`-2A+B+2(1-A)+1+B=0`
`-2A+B+2-2A+1+B=0`
`-4A+2B=-3` (equation 5)
Substitute the expressions of C and D in the fourth equation,
`8A+4B-8(1-A)+4(1+B)=0`
`8A+4B-8+8A+4+4B=0`
`16A+8B=4`
`4(4A+2B)=4`
`4A+2B=1` (equation 6)
Solve equation 5 and 6 to get the solutions of A and B,
Add equation 5 and 6,
`4B=-3+1`
`4B=-2`
`B=-2/4`
`B=-1/2`
Plug the value of B in the equation 5,
`-4A+2(-1/2)=-3`
`-4A-1=-3`
`-4A=-3+1`
`-4A=-2`
`A=1/2`
Now plug the values of A and B in the expressions of C and D,
`C=1-1/2`
`C=1/2`
`D=1+(-1/2)`
`D=1/2`
`:.x^3/((x+2)^2(x-2)^2)=1/(2(x+2))-1/(2(x+2)^2)+1/(2(x-2))+1/(2(x-2)^2)`
` `
No comments:
Post a Comment