Solve the differential equation $y’’ + 2y’ + 5y = \sin 2t$ using Laplace transform when $y=0$ and $y’ = 0$ when $t=0$
I've done the transform part and got a function –
$ y(S) = \frac{2}{ (s^2 +4)(s^2 + 2s + 5) } = \frac{A}{s^2 +4} + \frac{Bs+C}{ s^2 + 2s + 5} $
I know I need to complete the square later on to completely solve it but I’m having troubles with partial fractions in determining A , B and C.
From $ 2 = A(s^2 + 2s +5) + (Bs + C)(s^2 + 4) $
I am trying to find A first, then subsequently, substitute in the A value and the corresponding value to find B and C later on.
I can only find A if I make B and C = 0 . In other words, I know I have to make $(Bs + C)(s^2 + 4) = 0 $
But I’m stuck thinking of a value for s because of the power 2. I cannot sub in a negative 2 for example, as it will become 8 and not $ -2^2 + 4 = 0$ have I gotten something wrong or is there another way to do it ? Preferably I prefer doing it this way.
Best Answer
You will need $$\frac{2}{(s^2+4)(s^2+2s+5)}=\frac{As+B}{s^2+4}+\frac{Cs+D}{s^2+2s+5}\ .$$ Multiply out: $$2=(As+B)(s^2+2s+5)+(Cs+D)(s^2+4)\ ;\tag{$*$}$$ equate coefficients, $$A+C=0\ ,\quad 2A+B+D=0\ ,\quad 5A+2B+4C=0\ ,\quad 5B+4D=2\ ;$$ solve, which I leave up to you. When the denominators are quadratic there are typically no really good short cuts. You could try in $(*)$ substituting $s=2i$ to give $$2=(2Ai+B)(1+4i)$$ then dividing by $1+4i$ and equating real and imaginary parts, but this may well be more trouble than it's worth.