tgt

## 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
 $\dfrac{{15/4}}{{x - 1}} + \dfrac{{3/2}}{{{{(x - 1)}^2}}} + \dfrac{{ - 7}}{{x - 2}} + \dfrac{{13/4}}{{x - 3}}$