Express the term in partial fraction

Question

Came across this assignment question to express the term below in partial fraction

$$Y(S)=\frac{3e^{-4s}+1}{(s-3)(s+5)}$$

The proposed solution is to express it in partial fraction

$$Y(S)=\frac{3e^{-4s}+1}{(s-3)(s+5)}=\frac{A}{s-3}+\frac{B}{s-5}$$

by equating $s=3$ and $s=5$, obtained the constant for A and B as

$$A=\frac{3e^{-12}+1}{8}$$ and $$B=-\frac{3e^{20}+1}{8}$$

I don't understand as i don't seem to match the RHS equation as the same as original LHS equation after substitute back the value of A and B. Is this answer correct?

if it is not then it should be done as comparing the coefficient right?

by letting $$A(s+5)+B(s-3)=3e^{-4s}+1+0s$$

so getting A+B=0 and $$5A-3B=3e^{-4s}+1$$

getting A and B as $$A=\frac{3e^{-4s}+1}{8}$$ and $$B=-\frac{3e^{-4s}+1}{8}$$ and the RHS equation will be same as the LHS equation.

Answer 1

The proposed solution is not correct since $A$ and $B$ will not be constants and the denominator of the second fraction should be $s+5$.

Your solution is correct, indeed we have

$$\frac{3e^{-4s}+1}{(s-3)(s+5)}=\left(3e^{-4s}+1\right)\left(\frac{1}{(s-3)(s+5)}\right)$$ $$=\left(3e^{-4s}+1\right)\left(\frac{1}{8(s-3)}-\frac{1}{8(s+5)}\right)$$

Answer 2

Your second method is correct but the first one is incorrect because you've not considered the $e^{-4s}$ term in it.

It'd have worked good if it were of the form $\dfrac{as+b}{(s-c)(s-d)}$.

Here you need to let,

$$\dfrac{3e^{-4s}+1}{(s-3)(s+5)}=\dfrac{Ae^{-4s}+k}{s-3}+\frac{Be^{-4s}+h}{s-5}$$.

Now letting $s=3 \Rightarrow Ae^{-12}+k = \dfrac{3e^{-12}+1}{8} \Rightarrow A=\dfrac38,k=\dfrac18$

Similarly $s=-5 \Rightarrow Be^{20}+h = \dfrac{3e^{20}+1}{-8} \Rightarrow B=-\dfrac38,h=-\dfrac18$.

So,

$$\dfrac{3e^{-4s}+1}{(s-3)(s+5)}=\dfrac{3e^{-4s}+1}{8(s-3)}-\frac{3e^{-4s}+1}{8(s-5)}$$.

Now this is as expected.