I am to solve the following differential equation using laplace transforms $y''+y'-6y=15u(x-2)$ with the intial conditions $y(0)=1$ and $y'(0)=2$
So far I have calcualted the following
$$L(y'')+L(y')-L(6y)=L(15u(x-2))$$ $$s^2L(y)-sy(0)-y'(0)+sL(y)-y(0)-6L(y)= 15(\frac{e^{-2s}}{s})$$ $$(s^2+s-6)L(y)-(1+s)y(0)-y'(0)= 15(\frac{e^{-2s}}{s})$$ $$(s^2+s-6)L(y)-(1+s)-2= 15(\frac{e^{-2s}}{s})$$ $$L(y)=\frac{15(\frac{e^{-2s}}{s})+3+s}{s^2+s-6}$$
I am unsure of where to go from here to solve the Differential equation
$$L(y)=\frac{15(\frac{e^{-2s}}{s})+3+s}{s^2+s-6}=\frac{15(\frac{e^{-2s}}{s})+3+s}{(s+3)(s-2)}$$ Decompose the fractions: $$L(y)=15e^{-2s}\dfrac{1}{s(s+3)(s-2)}+\dfrac 1 {s-2}$$ $$L(y)=15e^{-2s}\left ( -\dfrac{1}{6s}+\dfrac {1}{15(s+3)}+\dfrac {1} {10(s-2)} \right)+\dfrac 1 {s-2}$$ $$L(y)=e^{-2s}\left ( -\dfrac{5}{2s}+\dfrac {1}{s+3}+\dfrac {3} {2(s-2)} \right)+\dfrac 1 {s-2}$$ Apply inverse Laplace transform now.