Laplace transform by partial equation

[nur aiman]

y" + 3y' + 4y = 6sin wt where y(0) = 4, y'(0) = 5

i have done halfway working. Is it correct the answer of differential equation? If correct, can u help for me to continue solving the question on getting the laplace transform.

 

[HallsofIvy]

I didn't check all of your calculations but it appears that the last formula you have is \(\displaystyle \frac{6\omega+ (s^2+ \omega^2)(4s+ 17)}{(s^2+ \omega^2)(s^2+ 3s+ 4)}\).

The denominator has the two factors \(\displaystyle s^2+ \omega^2\) and \(\displaystyle s^2+ 3s+ 4\). Neither of those is "reducible" to linear factors. We can write the first factor as the fraction \(\displaystyle \frac{As+ B}{s^2+ \omega^2}\). For the second we have to "complete the square". \(\displaystyle (3/2)^2= 9/4\) so adding and subtracting 9/4 we have \(\displaystyle s^2+ 3s+ 9/4- 9/4+ 4= (s+ 3/2)^2+ 7/4\). The second factor is \(\displaystyle \frac{Cs+ D}{(s+ 3/2)^2+ 7/4}\).