Решить систему ДУ с помощью преобразований Лапласса: x=x-2y+1, x(0)=0 y=-3x, y(0)=1

L{x'} = s*X(s) - x(0); L{x-2y+1} = X(s) - 2Y(s) +1/s;
L{y'} = s*Y(s) - y(0); L{-3x} = -3X(s);

s*X(s) = X(s) - 2Y(s) +1/s;
s*Y(s) - 1 = -3X(s);

s*X(s) = X(s) + 6X(s)/s -2/s +1/s;
Y(s) = -3X(s)/s + 1/s;

X(s)*(s-1-6/s) = - 1/s;
Y(s) = -3X(s)/s + 1/s;

X(s) = -1/(s^2-s-6)
Y(s) = -3X(s)/s + 1/s;

X(s)= -1/(s^2-s-6);
Y(s) = 3/(s^3-s^2-6s) + 1/s;

x(t) = L^-1{X(s)} = L^-1{-1/(s^2-s-6)} = L^-1{-1/(s^2-s+0.5^2 - 6.25)} = L^-1{-1/((s-0.5)^2 - 2.5^2)} = 0.4*sh(2.5*t)*exp(0.5*t)
Далее потребуется расписать:
3/(s^3-s^2-6s) = 3/s(s^2 - s - 6) = A/s + (Bs+C)/(s^2 - s - 6) = (A*s^2-A*s-A*6 +B*s^2+C*s)/s(s^2 - s - 6)
A+B = 0
-A + C = 0
-A*6 = 3
A = -0.5, B = 0.5, C = -0.5
3/(s^3-s^2-6s) = -0.5/s +0.5s/(s^2 - s - 6) - 0.5/(s^2 - s - 6) =
= -0.5/s - 0.5/((s-0.5)^2 - 2.5^2) + 0.5s/((s-0.5)^2 - 2.5^2) =
= -0.5/s - 0.5/((s-0.5)^2 - 2.5^2) + 0.5(s-0.5)/((s-0.5)^2 - 2.5^2) + 0.25/((s-0.5)^2 - 2.5^2) =
= -0.5/s - 0.25/((s-0.5)^2 - 2.5^2) + 0.5(s-0.5)/((s-0.5)^2 - 2.5^2)

y(t) = L^-1{Y(s)} = L^-1{-0.5/s - 0.25/((s-0.5)^2 - 2.5^2) + 0.5(s-0.5)/((s-0.5)^2 - 2.5^2 + 1/s} =
= L^-1{0.5/s - 0.25/((s-0.5)^2 - 2.5^2) + 0.5(s-0.5)/((s-0.5)^2 - 2.5^2} =
= 0.5 - 0.1*sh(2.5*t)*exp(0.5*t) + 0.5*ch(2.5*t)*exp(0.5*t)

x(t) = 0.4*sh(2.5*t)*exp(0.5*t)
y(t) = 0.5 - 0.1*sh(2.5*t)*exp(0.5*t) + 0.5*ch(2.5*t)*exp(0.5*t)