Question: Solve y''(t) - 2y'(t)-3y(t) = 0 where y(0) = 1 and y'(0) = 0.
L{y″(t)}−2⋅L{y′(t)}−3⋅L{y(t)}=L{0}
L{y″(t)}=s2⋅L{y(t)}−s⋅y(0)−y′(0)L{y″(t)}=s2⋅L{y(t)}−s⋅1−0L{y″(t)}=s2⋅L{y(t)}−s
−2⋅L{y′(t)}=−2[s⋅L{y(t)}−y(0)]−2⋅L{y′(t)}=−2[s⋅L{y(t)}−1]
s2⋅L{y(t)}−s−2[s⋅L{y(t)}−1]−3⋅L{y(t)}=L{0}L{y(t)}[s2−2⋅s−3]=s−2L{y(t)}[(s+1)⋅(s−3)]=s−2L{y(t)}=s−2(s+1)⋅(s−3)
s−2(s+1)⋅(s−3)=As+1+Bs−3s−2=A⋅(s−3)+B⋅(s+1)s=3:3−2=0+B⋅4⇒B=14s=−1:−1−2=A⋅(−4)+0⇒A=34
L{y(t)}=s−2(s+1)⋅(s−3)=As+1+Bs−3=34⋅(1s+1)+14⋅(1s−3)L{y(t)}=34⋅(1s+1)+14⋅(1s−3)
L−1{1s−a}=ea⋅t
\textcolor[rgb]{150,0,0}{ \begin{array}{rcl} y(t) &=& \dfrac{3}{4}\cdot e^{-t} + \dfrac{1}{4} \cdot e^{3\cdot t} \end{array} }
Check:
\begin{array}{lrcl} (1)&: y(0) &=& \dfrac{3}{4}+\dfrac{1}{4} = 1 \qquad \text{okay} \\\\ (2)&: y'(t) &=& -\dfrac{3}{4}\cdot e^{-t} + 3\cdot \dfrac{1}{4} \cdot e^{3\cdot t}} \\\\ &: y'(0) &=& -\dfrac{3}{4} + \dfrac{3}{4} = 0 \qquad \text{okay}\\ \end{array}\\\\\\ \underbrace{ \begin{array}{lrcl} &y''(t) &=& \dfrac{3}{4}\cdot e^{-t} + \dfrac{9}{4}\cdot e^{3\cdot t} \\\\ & -2\cdot y'(t) & =& \dfrac{6}{4}\cdot e^{-t} -\dfrac{6}{4}\cdot e^ {3\cdot t} \\\\ & -3\cdot y(t) & =& -\dfrac{9}{4}\cdot e^{-t} -\dfrac{3}{4}\cdot e^{3\cdot t} \end{array} }_{ y''(t)-2\cdot y'(t)-3\cdot y(t)= 0 \qquad \text{okay} }
.
