1. Pull-change is a mathematical concept, so any function that meets the definition of Laplace transform can be pulled-change, so the step of transfer function can be omitted.

2.(s？ +...)X0(s)-[ ... This step is based on the differential property of Lagrangian change. I don't know what you mean by directly carrying out Lagrangian transformation, (S? +...)X0(s)-[...] This step itself is a direct pull change. Just a combination of similar items, Ld? x0(t)/dt？ =s？ X0(s)-sX0(0)-X'0(0)，L5dx0(t)/dt=5sX0(s)-5X0(0)，L6x0(t)=6X0(s).

Maybe you mean that there is no -[(s+5)X0(0)+X'0(0)] in the example in the book. Because x0(0)=0 by default in the automatic control book, the item in -[(s+5)X0(0)+X'0(0)] is zero, so it is ok.

3。 It says -[(s+5)X0(0)+X'0(0)]=0, then the left =(s? +5s+6)X0(s), right =1/s. That is to say, (s? +5s+6)X0(s)= 1/s, so X0(s)= 1/[s*(s? +5s+6)]= 1/[s *(s+2)*(s+3)]

According to 3, the poles are 0, -2, -3. Now you should know how to do it.

5. He said it was a look-up table. You won't be given a form in the exam. You'd better write it down, just write something simple.

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