How reduce the steps of lengthy answer (partial fraction decomposition)

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PA3040D

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Dear expert, I have answered the exam question attached below, but I believe that the answer is correct. However, the process is lengthy and time-consuming. Could you please assist me in simplifying the steps that reducing the time required?

Find the inverse Laplace of


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As my experience tells me, there is no shortcut for doing partial fraction decomposition by hand. But with time, you will be quicker.
 
PA3040D said:
Dear expert, I have answered the exam question attached below, but I believe that the answer is correct. However, the process is lengthy and time-consuming. Could you please assist me in simplifying the steps that reducing the time required?
Click to expand...
There is, in fact, a shortcut for finding partial fractions when the denominators are distinct linear factors:
1716986516973.png
(By the way, technically you omitted the denominator on the first line, but the ultimate equation is correct.)

What you do is to set ss to each of the zeros of the denominators, one at a time:

s=1s=1: (1)2−6(1)+10=A(1−2)(1−1)+B(1−3)(1−1)+C(1−3)(1−2)  ⟹  5=0+0+2C  ⟹  C=5/2(1)^2-6(1)+10=A(1-2){\color{Red}(1-1)}+B(1-3){\color{Red}(1-1)}+C(1-3)(1-2)\implies 5={\color{Red}0}+{\color{Red}0}+2C\implies C=5/2

s=2s=2: (2)2−6(2)+10=A(2−2)(2−1)+B(2−3)(2−1)+C(2−3)(2−2)  ⟹  2=0−B+0  ⟹  B=−2(2)^2-6(2)+10=A{\color{Red}(2-2)}(2-1)+B(2-3)(2-1)+C(2-3){\color{Red}(2-2)}\implies 2={\color{Red}0}-B+{\color{Red}0}\implies B=-2

s=3s=3: (3)2−6(3)+10=A(3−2)(3−1)+B(3−3)(3−1)+C(3−3)(3−2)  ⟹  1=2A+0+0  ⟹  A=1/2(3)^2-6(3)+10=A(3-2)(3-1)+B{\color{Red}(3-3)}(3-1)+C{\color{Red}(3-3)}(3-2)\implies 1=2A+{\color{Red}0}+{\color{Red}0}\implies A=1/2
 
Dr.Peterson said:
There is, in fact, a shortcut for finding partial fractions when the denominators are distinct linear factors:
View attachment 38013
(By the way, technically you omitted the denominator on the first line, but the ultimate equation is correct.)

What you do is to set ss to each of the zeros of the denominators, one at a time:

s=1s=1: (1)2−6(1)+10=A(1−2)(1−1)+B(1−3)(1−1)+C(1−3)(1−2)  ⟹  5=0+0+2C  ⟹  C=5/2(1)^2-6(1)+10=A(1-2){\color{Red}(1-1)}+B(1-3){\color{Red}(1-1)}+C(1-3)(1-2)\implies 5={\color{Red}0}+{\color{Red}0}+2C\implies C=5/2

s=2s=2: (2)2−6(2)+10=A(2−2)(2−1)+B(2−3)(2−1)+C(2−3)(2−2)  ⟹  2=0−B+0  ⟹  B=−2(2)^2-6(2)+10=A{\color{Red}(2-2)}(2-1)+B(2-3)(2-1)+C(2-3){\color{Red}(2-2)}\implies 2={\color{Red}0}-B+{\color{Red}0}\implies B=-2

s=3s=3: (3)2−6(3)+10=A(3−2)(3−1)+B(3−3)(3−1)+C(3−3)(3−2)  ⟹  1=2A+0+0  ⟹  A=1/2(3)^2-6(3)+10=A(3-2)(3-1)+B{\color{Red}(3-3)}(3-1)+C{\color{Red}(3-3)}(3-2)\implies 1=2A+{\color{Red}0}+{\color{Red}0}\implies A=1/2
Click to expand...
This is fine ...Grate thanks
 
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