Решить по Лапласу y’’-2y’-3y=2t y(0)=1, y’(0)=1

Решите задачу:

$p^2Y(p)-py(0)-y'(0)-2(pY(p)-y(0))-3Y(p)=\dfrac{2}{p^2}\\ Y(p)*(p^2-2p-3)=\dfrac{2}{p^2}+p-1\\ Y(p)=\dfrac{2}{p^2(p^2-2p-3)}+\dfrac{p-1}{(p-3)(p+1)}\\ Y(p)=\dfrac{1}{18(p-3)}-\dfrac{1}{2(p+1)}+\dfrac{4p-6}{9p^2}+\dfrac{1}{2(p-3)}+\dfrac{1}{2(p+1)}\\ Y(p)=\dfrac{5}{9(p-3)}+\dfrac{4p-6}{9p^2}\\ y(t)=\dfrac{5e^{3t}}{9}+\dfrac{4}{9}-\dfrac{2}{3} t$