Finding the inverse laplace transform

[Ry122]

I'm attempting to find the inverse laplace transform of

$$\frac{25}{(1-s)^2*(4+s^2)}$$

I get to this point but can't get the values of A B and C when equating coefficients.

25/(s-1)=A(4+s^2)+(Bs+C)(s-1)


Also for a separate question: (s+1)/(5-4s+s^2)

How do you find the inverse laplace transform when you can't perform partial fraction expansion?

 

[RoshanBBQ]

You are not expanding the first one right, because you are ignoring the effect of repeated roots.
$$\frac{25}{(1-s)^2(s^2+4)}=\frac{A}{1-s}+\frac{B}{(1-s)^2}+\frac{Cs+D}{s^2+4}$$

As for the other, my hint is to complete the square in the denominator. These, if I remember correctly, turn out to be a sine plus a cosine (or sometimes just a cosine or a sine).

 

[Ry122]

when you complete the square for the 2nd one you get 1+(s-2)^2
so would you just have one residual and that term in the denominator and solve the inverse laplace from there?

 

[RoshanBBQ]

when you complete the square for the 2nd one you get 1+(s-2)^2
so would you just have one residual and that term in the denominator and solve the inverse laplace from there?

Once you complete the square, you should already be very close to a form ready for an inverse Laplace.

$$e^{-at}cos(bt)u(t) \leftrightarrow \frac{s+a}{(s+a)^2+b^2}$$
$$e^{-at}sin(bt)u(t) \leftrightarrow \frac{b}{(s+a)^2+b^2}$$
So in general, if you have
$$\frac{s+U}{(s+a)^2+b^2}$$
You can do the following manipulations:
$$\frac{s+a - a+ U}{(s+a)^2+b^2}=\frac{s+a}{(s+a)^2+b^2}+\frac{U-a}{(s+a)^2+b^2}=\frac{s+a}{(s+a)^2+b^2}+\frac{U-a}{b}\frac{b}{(s+a)^2+b^2} \leftrightarrow e^{-at}cos(bt)u(t)+\frac{U-a}{b}e^{-at}sin(bt)u(t)$$
where U is a constant. But I see an issue, your a is -2, resulting in e^(2t). Are you doing right-handed Laplace transforms only?