L[x″]=s2X(s)−sx(t0)−x′(t0)=s2X(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′]=sX(s)−x(t0)=sX(s)−1{\displaystyle {\mathcal {L}}[x']=sX(s)-x(t_{0})=sX(s)-1}
Reemplazando en la ecuación
L[x″+4x′+3x]=L[e−t]{\displaystyle {\mathcal {L}}[x''+4x'+3x]={\mathcal {L}}[e^{-t}]}
s2X(s)−s−1+4sX(s)−4+3X(s)=1s+1{\displaystyle s^{2}X(s)-s-1+4sX(s)-4+3X(s)={\frac {1}{s+1}}}
X(s)(s2+4s+3)−s−5=1s+1{\displaystyle X(s)(s^{2}+4s+3)-s-5={\frac {1}{s+1}}}
X(s)(s2+4s+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)(s2+4s+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)=A2(s+1)2+A1s+1+Bs+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 A2{\displaystyle A_{2}} y de B{\displaystyle B} se pueden obtener directamente:
A2=[1+(s+5)(s+1)s+3]s=−1=12{\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=−34{\displaystyle B=\left[{\frac {1+(s+5)(s+1)}{(s+1)^{2}}}\right]_{s=-3}=-{\frac {3}{4}}}
Para obtener el valor de A1{\displaystyle A_{1}} pueden utilizarse estos valores obtenidos y resolver el sistema, o puede utilizarse la derivada:
A1=dds[1+(s+5)(s+1)s+3]s=−1=[1+3(s+3)2]s=−1=74{\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)=12(s+1)2+74s+1+34s+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)=12te−t+74et−34e−3t{\displaystyle {\mathcal {L}}^{-1}[X(s)]=x(t)={\frac {1}{2}}te^{-t}+{\frac {7}{4}}e^{t}-{\frac {3}{4}}e^{-3t}}
![{\displaystyle {\mathcal {L}}[x'']=s^{2}X(s)-sx(t_{0})-x'(t_{0})=s^{2}X(s)-s-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/883d466697ea65e42384670a59b0a39f76e4dd7c)
![{\displaystyle {\mathcal {L}}[x']=sX(s)-x(t_{0})=sX(s)-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5a1dcf2a8bd8d9598d8e196afad444b1b81c85c)
![{\displaystyle {\mathcal {L}}[x''+4x'+3x]={\mathcal {L}}[e^{-t}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebd21e5ff74995ed7430deb8f2f5820bec9cd6b0)
y de 
se pueden obtener directamente:
![{\displaystyle A_{2}=\left[{\frac {1+(s+5)(s+1)}{s+3}}\right]_{s=-1}={\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf7de49d3799563e4c87203c605cc78db0fc4a73)
![{\displaystyle B=\left[{\frac {1+(s+5)(s+1)}{(s+1)^{2}}}\right]_{s=-3}=-{\frac {3}{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bfecec28e46f0d808eb8b4a4616375434138d13)
pueden utilizarse estos valores obtenidos y resolver el sistema, o puede utilizarse la derivada:
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8deacdfd983664423b7960f09561f0f71bbcdd68)
![{\displaystyle {\mathcal {L}}^{-1}[X(s)]=x(t)={\frac {1}{2}}te^{-t}+{\frac {7}{4}}e^{t}-{\frac {3}{4}}e^{-3t}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6def14b955ade38a3e2fd255c34606698ad24447)
