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X ′ ( t ) = [ 2 − 5 1 − 2 ] X + [ − cos t sin t ] {\displaystyle X'(t)={\begin{bmatrix}2&-5\\1&-2\end{bmatrix}}\,X+{\begin{bmatrix}-\cos {t}\\\sin {t}\end{bmatrix}}}
Eigenvalues: λ ∈ { − i , i } {\displaystyle \lambda \in \{-i,i\}}
X h ( t ) = c 1 ( v → 1 cos t − v → 2 sin t ) + c 2 ( v → 2 cos t + v → 1 sin t ) {\displaystyle X_{h}(t)=c_{1}\,\left({\vec {v}}_{1}\,\cos {t}-{\vec {v}}_{2}\,\sin {t}\right)+c_{2}\,\left({\vec {v}}_{2}\,\cos {t}+{\vec {v}}_{1}\,\sin {t}\right)}
Guess for X p ( t ) {\displaystyle X_{p}(t)} :
X ′ ( t ) = [ 0 1 1 0 ] X + [ t − 1 ] X ( 0 ) = [ 2 1 ] {\displaystyle X'(t)={\begin{bmatrix}0&1\\1&0\end{bmatrix}}\,X+{\begin{bmatrix}t\\-1\end{bmatrix}}\quad \quad X(0)={\begin{bmatrix}2\\1\end{bmatrix}}}
{ L { x ′ } = L { y } + L { t } L { y ′ } = L { x } + L { − 1 } {\displaystyle {\begin{cases}{\mathcal {L}}\left\{x'\right\}={\mathcal {L}}\left\{y\right\}+{\mathcal {L}}\left\{t\right\}\\{\mathcal {L}}\left\{y'\right\}={\mathcal {L}}\left\{x\right\}+{\mathcal {L}}\left\{-1\right\}\end{cases}}}
{ s L { x } − x ( 0 ) = L { y } + 1 s 2 s L { y } − y ( 0 ) = L { x } + 1 s {\displaystyle {\begin{cases}s\,{\mathcal {L}}\left\{x\right\}-x(0)={\mathcal {L}}\left\{y\right\}+{\frac {1}{s^{2}}}\\s\,{\mathcal {L}}\left\{y\right\}-y(0)={\mathcal {L}}\left\{x\right\}+{\frac {1}{s}}\end{cases}}}
{ s L { x } − 2 = L { y } + 1 s 2 s L { y } − 1 = L { x } + 1 s {\displaystyle {\begin{cases}s\,{\mathcal {L}}\left\{x\right\}-2={\mathcal {L}}\left\{y\right\}+{\frac {1}{s^{2}}}\\s\,{\mathcal {L}}\left\{y\right\}-1={\mathcal {L}}\left\{x\right\}+{\frac {1}{s}}\end{cases}}}
L { x } = s L { y } − 1 + 1 s {\displaystyle {\mathcal {L}}\left\{x\right\}=s\,{\mathcal {L}}\left\{y\right\}-1+{\frac {1}{s}}}
L { y } = 1 s 2 ( s 2 − 1 ) + 1 s 2 − 1 + s s 2 − 1 {\displaystyle {\mathcal {L}}\left\{y\right\}={\frac {1}{s^{2}\,\left(s^{2}-1\right)}}+{\frac {1}{s^{2}-1}}+{\frac {s}{s^{2}-1}}}
L { x } = s ( 1 s 2 ( s 2 − 1 ) + 1 s 2 − 1 + s s 2 − 1 ) − 1 + 1 s {\displaystyle {\mathcal {L}}\left\{x\right\}=s\,\left({\frac {1}{s^{2}\,\left(s^{2}-1\right)}}+{\frac {1}{s^{2}-1}}+{\frac {s}{s^{2}-1}}\right)-1+{\frac {1}{s}}}
Take inverse Laplace transforms
Using matrix notation,
X ′ ( t ) = [ 2 − 1 3 − 2 ] X ( t ) + [ 1 − 1 ] e t X ( 0 ) = [ 3 0 ] {\displaystyle X'(t)={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,X(t)+{\begin{bmatrix}1\\-1\end{bmatrix}}\,\mathrm {e} ^{t}\quad \quad X(0)={\begin{bmatrix}3\\0\end{bmatrix}}}
L { X ′ } = [ 2 − 1 3 − 2 ] L { X } + [ 1 , − 1 ] L { e t } s L { X } − X ( 0 ) = [ 2 − 1 3 − 2 ] L { X } + [ 1 − 1 ] ( 1 s − 1 ) ( s I − [ 2 − 1 3 − 2 ] ) L { X } = X ( 0 ) + 1 s − 1 [ 1 − 1 ] L { X } = [ s − 2 1 − 3 s + 2 ] − 1 ( X ( 0 ) + 1 s − 1 [ 1 − 1 ] ) L { X } = 1 s 2 − 1 [ s + 2 − 1 3 s − 2 ] [ 3 + 1 s − 1 − 1 s − 1 ] {\displaystyle {\begin{aligned}{\mathcal {L}}\left\{X'\right\}&={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,{\mathcal {L}}\left\{X\right\}+{\begin{bmatrix}1,-1\end{bmatrix}}\,{\mathcal {L}}\left\{\mathrm {e} ^{t}\right\}\\s\,{\mathcal {L}}\left\{X\right\}-X(0)&={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,{\mathcal {L}}\left\{X\right\}+{\begin{bmatrix}1\\-1\end{bmatrix}}\,\left({\frac {1}{s-1}}\right)\\\left(s\,I-{\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\right)\,{\mathcal {L}}\left\{X\right\}&=X(0)+{\frac {1}{s-1}}\,{\begin{bmatrix}1\\-1\end{bmatrix}}\\{\mathcal {L}}\left\{X\right\}&={\begin{bmatrix}s-2&1\\-3&s+2\end{bmatrix}}^{-1}\,\left(X(0)+{\frac {1}{s-1}}\,{\begin{bmatrix}1\\-1\end{bmatrix}}\right)\\{\mathcal {L}}\left\{X\right\}&={\frac {1}{s^{2}-1}}\,{\begin{bmatrix}s+2&-1\\3&s-2\end{bmatrix}}\,{\begin{bmatrix}3+{\frac {1}{s-1}}\\-{\frac {1}{s-1}}\end{bmatrix}}\end{aligned}}}
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