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Monday, 4 August 2014

CHAPTER 11- Worked Out Examples

Example: 1     

Expand the following using partial dfractions:
(a) \dfrac{{{x^2} - 3x - 4}}{{{x^3} - 6{x^2} + 11x - 6}}
(b) \dfrac{{{x^2} + x + 1}}{{{{(x - 1)}^2}(x - 2)(x - 3)}}
Solution: 1-(a)

The denominator can be factorised:
{x^3} - 6{x^2} + 11x - 6 = (x - 1)(x - 2)(x - 3)
The partial dfraction expansion is:
\dfrac{{(x + 1)(x - 4)}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{x - 2}} + \dfrac{C}{{x - 3}}
Cross-multiplying, we obtain
(x + 1)(x - 4) = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)
Put x = 1 \Rightarrow\,\,\,\, A = \dfrac{{(2) \times ( - 3)}}{{( - 1) \times ( - 2)}} =  - 3
Put x = 2 \Rightarrow\,\,\,\, B = \dfrac{{(3) \times ( - 2)}}{{(1) \times ( - 1)}} = 6
Put x = 3 \Rightarrow\,\,\,\, C = \dfrac{{(4) \times ( - 1)}}{{(2) \times (1)}} =  - 2
Thus, the partial dfraction expansion is
\dfrac{{ - 3}}{{x - 1}} + \dfrac{6}{{x - 2}} + \dfrac{{ - 2}}{{x - 3}}
Solution: 1-(b)

As discussed previously, the partial dfraction expansion of this expression would be of the form
\dfrac{{{x^2} + x + 1}}{{{{(x - 1)}^2}(x - 2)(x - 3)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{x - 2}} + \dfrac{D}{{x - 3}}
Cross-multiplying, we obtain
{x^2} + x + 1 = A(x - 1)(x - 2)(x - 3) + B(x - 2)(x - 3) + C{(x - 1)^2}(x - 3) + D{(x - 1)^2}(x - 2)
Put x = 1 \Rightarrow\,\,\,\, B = \dfrac{{1 + 1 + 1}}{{( - 1) \times ( - 2)}} = \dfrac{3}{2}
x = 2 \Rightarrow\,\,\,\, C = \dfrac{{4 + 2 + 1}}{{{1^2} \times ( - 1)}} =  - 7
x = 3 \Rightarrow\,\,\,\, D = \dfrac{{9 + 3 + 1}}{{{2^2} \times (1)}} = \dfrac{{13}}{4}
To obtain A, we compare the coefficients of {x^3} on both sides. Thus,
0 = A + C + D
 \Rightarrow\,\,\,\, A =  - (C + D)
\,\,\,\,\, =  - \left( { - 7 + \dfrac{{13}}{4}} \right)
\,\,\,\,\, = \dfrac{{15}}{4}
The required partial dfraction expansion is
\dfrac{{15/4}}{{x - 1}} + \dfrac{{3/2}}{{{{(x - 1)}^2}}} + \dfrac{{ - 7}}{{x - 2}} + \dfrac{{13/4}}{{x - 3}}
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