Last night I attempted and successfully finished (with the help of stackexchange) the first part to this question on laplace transformations:
https://math.stackexchange.com/questions/468596/laplace-question-help-needed
The second part to this question is:
Here are my working/calculations so far:
I know I am going wrong somewhere but I don't know where. When solving the last two equations (shown) simultaneously I am getting different answers for C.
Any help on this is really appreciated! thank you :)
You appear to be working with (using a conventional general form of partial fractions) $$\frac{1}{s^2(s^2+1)}+\frac{s-2}{s^2+1}=\frac{As+B}{s^2}+\frac{Cs+D}{s^2+1}$$ Multiply through by $s^2(s^2+1)$ to clear fractions$$1+s^2(s-2)=(As+B)(s^2+1)+(Cs+D)s^2$$ We then set $s=0$ to obtain $B=1$ and setting $s=i$ gives $$3-i=-Ci-D$$ We can then equate the coefficients of $s^3$ on each side to obtain $A$.
Another route would be to work only with the first term on the left (since the second term on the left is in the correct form anyway), and note that this is an expression in $t=s^2$ with $$\frac 1{t(t+1)}=\frac At+\frac B{t+1}=\frac 1t-\frac 1{t+1}$$ where we use conventional partial fractions technique (or prior knowledge of a common form) to obtain the expression.
Using the first method we obtain $B=1, C=1, D=-3$ from the given equations. Equating coefficients of $s^3$ gives $s^3=As^3+Cs^3$ and since $C=1$ we have $A=0$ giving $$\frac{1}{s^2}+\frac{s-3}{s^2+1}=\frac{1}{s^2}+\frac{s}{s^2+1}-\frac{3}{s^2+1}$$
Using the second method ($t=s^2$) we get $$\frac 1t-\frac 1{t+1}+\frac{s-2}{s^2+1}=\frac 1{s^2}-\frac 1{s^2+1}+\frac{s-2}{s^2+1}=\frac1{s^2}+\frac s{s^2+1}-\frac3{s^2+1}$$