L [ x ″ ] = s 2 X ( s ) − s x ( t 0 ) − x ′ ( t 0 ) = s 2 X ( s ) − s − 1 {\displaystyle {\mathcal {L}}[x'']=s^{2}X(s)-sx(t_{0})-x'(t_{0})=s^{2}X(s)-s-1}
L [ x ′ ] = s X ( s ) − x ( t 0 ) = s X ( s ) − 1 {\displaystyle {\mathcal {L}}[x']=sX(s)-x(t_{0})=sX(s)-1}
Reemplazando en la ecuación
L [ x ″ + 4 x ′ + 3 x ] = L [ e − t ] {\displaystyle {\mathcal {L}}[x''+4x'+3x]={\mathcal {L}}[e^{-t}]}
s 2 X ( s ) − s − 1 + 4 s X ( s ) − 4 + 3 X ( s ) = 1 s + 1 {\displaystyle s^{2}X(s)-s-1+4sX(s)-4+3X(s)={\frac {1}{s+1}}}
X ( s ) ( s 2 + 4 s + 3 ) − s − 5 = 1 s + 1 {\displaystyle X(s)(s^{2}+4s+3)-s-5={\frac {1}{s+1}}}
X ( s ) ( s 2 + 4 s + 3 ) = 1 + ( s + 5 ) ( s + 1 ) s + 1 {\displaystyle X(s)(s^{2}+4s+3)={\frac {1+(s+5)(s+1)}{s+1}}}
X ( s ) = 1 + ( s + 5 ) ( s + 1 ) ( s + 1 ) ( s 2 + 4 s + 3 ) {\displaystyle X(s)={\frac {1+(s+5)(s+1)}{(s+1)(s^{2}+4s+3)}}}
X ( s ) = 1 + ( s + 5 ) ( s + 1 ) ( s + 1 ) ( s + 1 ) ( s + 3 ) {\displaystyle X(s)={\frac {1+(s+5)(s+1)}{(s+1)(s+1)(s+3)}}}
X ( s ) = 1 + ( s + 5 ) ( s + 1 ) ( s + 1 ) 2 ( s + 3 ) = A 2 ( s + 1 ) 2 + A 1 s + 1 + B s + 3 {\displaystyle X(s)={\frac {1+(s+5)(s+1)}{(s+1)^{2}(s+3)}}={\frac {A_{2}}{(s+1)^{2}}}+{\frac {A_{1}}{s+1}}+{\frac {B}{s+3}}}
Los valores de A 2 {\displaystyle A_{2}} y de B {\displaystyle B} se pueden obtener directamente:
A 2 = [ 1 + ( s + 5 ) ( s + 1 ) s + 3 ] s = − 1 = 1 2 {\displaystyle A_{2}=\left[{\frac {1+(s+5)(s+1)}{s+3}}\right]_{s=-1}={\frac {1}{2}}}
B = [ 1 + ( s + 5 ) ( s + 1 ) ( s + 1 ) 2 ] s = − 3 = − 3 4 {\displaystyle B=\left[{\frac {1+(s+5)(s+1)}{(s+1)^{2}}}\right]_{s=-3}=-{\frac {3}{4}}}
Para obtener el valor de A 1 {\displaystyle A_{1}} pueden utilizarse estos valores obtenidos y resolver el sistema, o puede utilizarse la derivada:
A 1 = d d s [ 1 + ( s + 5 ) ( s + 1 ) s + 3 ] s = − 1 = [ 1 + 3 ( s + 3 ) 2 ] s = − 1 = 7 4 {\displaystyle A_{1}={\frac {d}{ds}}\left[{\frac {1+(s+5)(s+1)}{s+3}}\right]_{s=-1}=\left[1+{\frac {3}{(s+3)^{2}}}\right]_{s=-1}={\frac {7}{4}}}
X ( s ) = 1 2 ( s + 1 ) 2 + 7 4 s + 1 + 3 4 s + 3 {\displaystyle X(s)={\frac {\frac {1}{2}}{(s+1)^{2}}}+{\frac {\frac {7}{4}}{s+1}}+{\frac {\frac {3}{4}}{s+3}}}
L − 1 [ X ( s ) ] = x ( t ) = 1 2 t e − t + 7 4 e t − 3 4 e − 3 t {\displaystyle {\mathcal {L}}^{-1}[X(s)]=x(t)={\frac {1}{2}}te^{-t}+{\frac {7}{4}}e^{t}-{\frac {3}{4}}e^{-3t}}