y''(t)+5y'(t)+4y(t)=exp(2t)
L(y'')=s²y(s)
L(5y')=5sy(s)
L(4y)=4y(s)
L(exp(2t))=1/(s-2)
y(s)(s²+5s+4)=1/(s-2)
y(s)((s+4)(s+1))=1/(s-2)
y(s)=1/((s-2)(s+4)(s+1))
Rozkładam na ułamki proste
1/((s-2)(s+4)(s+1)) ≡ A/(s-2) + B/(s+1) + C(s+4)
A((s+1)(s+4)) + B((s+4)(s-2)) + C((s+1)(s-2)) ≡ 1
A(s²+5s+4) +B(s²+2s-8) +C(s²-s-2) ≡ 1
s²: A+B+C=0 => C=-(A+B)
s: 5A+2B-C=0 =>5A+2B + A + B =0 => 6A+3B=0 => 2A=-B
C=A
1: 4A-8B-2C=1 => 4A+16A-2A=1 => A=1/18, B=-1/9
Obliczam transformatę odwrotną:
y(t)= (1/18)*exp(2t)-(1/9)*exp(-t)+(1/18)*exp(-4t)
