制御と振動の数学/第一類/連立微分方程式の解法/例題による考察/同次微分方程式

例104 {\displaystyle \quad }
を解け.
これは普通に解けばよい. x ⊐ X , y ⊐ Y {\displaystyle x\sqsupset X,y\sqsupset Y} とおいて式 (5.1) をLaplace 変換すれば
この原像を求めると,
を得る. ♢ {\displaystyle \diamondsuit }


例105 {\displaystyle \quad }
次の連立微分方程式を解け.
解答例
X ⊏ x , Y ⊏ y {\displaystyle X\sqsubset x,Y\sqsubset y} とおいて,与方程式を Laplace 変換すると, ( s + 1 − 3 2 s − 4 ) ( X Y ) = ( α β ) {\displaystyle {\begin{pmatrix}s+1&-3\\2&s-4\end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}\alpha \\\beta \end{pmatrix}}} ∴ ( X Y ) = ( s + 1 − 3 2 s − 4 ) − 1 ( α β ) = 1 s 2 − 3 s + 2 ( s − 4 3 − 2 s + 1 ) ( α β ) = 1 ( s − 1 ) ( s − 2 ) ( α ( s − 4 ) + 3 β − 2 α + β ( s + 1 ) ) {\displaystyle \therefore {\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}s+1&-3\\2&s-4\end{pmatrix}}^{-1}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}={\frac {1}{s^{2}-3s+2}}{\begin{pmatrix}s-4&3\\-2&s+1\end{pmatrix}}{\begin{pmatrix}\alpha \\\beta \end{pmatrix}}={\frac {1}{(s-1)(s-2)}}{\begin{pmatrix}\alpha (s-4)+3\beta \\-2\alpha +\beta (s+1)\end{pmatrix}}} α s − 4 α ( s − 1 ) ( s − 2 ) = 3 α s − 1 − 2 α s − 2 {\displaystyle {\frac {\alpha s-4\alpha }{(s-1)(s-2)}}={\frac {3\alpha }{s-1}}-{\frac {2\alpha }{s-2}}} ...1 3 β ( s − 1 ) ( s − 2 ) = − 3 β s − 1 + 3 β s − 2 {\displaystyle {\frac {3\beta }{(s-1)(s-2)}}={\frac {-3\beta }{s-1}}+{\frac {3\beta }{s-2}}} ...2 12の原像は, x = 3 α e t − 2 α e 2 t − 3 β e t + 3 β e 2 t {\displaystyle x=3\alpha e^{t}-2\alpha e^{2t}-3\beta e^{t}+3\beta e^{2t}} ∴ x = α ( 3 e t − 2 e 2 t ) + 3 ( − e t + e 2 t ) β . {\displaystyle \therefore x=\alpha (3e^{t}-2e^{2t})+3(-e^{t}+e^{2t})\beta .} 同様に − 2 α ( s − 1 ) ( s − 2 ) = 2 α ( 1 s − 1 − 1 s − 2 ) {\displaystyle {\frac {-2\alpha }{(s-1)(s-2)}}=2\alpha \left({\frac {1}{s-1}}-{\frac {1}{s-2}}\right)} ...3 β ( s + 1 ) ( s − 1 ) ( s − 2 ) = − 2 β s − 1 + 3 β s − 2 {\displaystyle {\frac {\beta (s+1)}{(s-1)(s-2)}}={\frac {-2\beta }{s-1}}+{\frac {3\beta }{s-2}}} ...4 34の原像は, y = 2 α e t − 2 α e 2 t − 2 β e t + 3 β e 2 t {\displaystyle y=2\alpha e^{t}-2\alpha e^{2t}-2\beta e^{t}+3\beta e^{2t}} ∴ y = 2 ( e t − e 2 t ) α + ( − 2 e t + 3 e 2 t ) β {\displaystyle \therefore y=2(e^{t}-e^{2t})\alpha +(-2e^{t}+3e^{2t})\beta } ♢ {\displaystyle \diamondsuit }

例106 {\displaystyle \quad }
次の連立微分方程式を解け.
解答例
X ⊏ x , Y ⊏ y {\displaystyle X\sqsubset x,Y\sqsubset y} とおいて,与方程式を Laplace 変換すると, { s X − α X − β Y = β s − α s Y − 1 + β X − α Y = 0 {\displaystyle {\begin{cases}sX-\alpha X-\beta Y={\frac {\beta }{s-\alpha }}\\sY-1+\beta X-\alpha Y=0\end{cases}}}
∴ ( s − α − β β s − α ) ( X Y ) = ( β s − α 1 ) {\displaystyle \therefore {\begin{pmatrix}s-\alpha &-\beta \\\beta &s-\alpha \end{pmatrix}}{\begin{pmatrix}X\\Y\end{pmatrix}}={\begin{pmatrix}{\frac {\beta }{s-\alpha }}\\1\end{pmatrix}}} ∴ ( X Y ) = 1 ( s − α ) 2 + β 2 ( s − α β − β s − α ) ( β s − α 1 ) {\displaystyle \therefore {\begin{pmatrix}X\\Y\end{pmatrix}}={\frac {1}{(s-\alpha )^{2}+\beta ^{2}}}{\begin{pmatrix}s-\alpha &\beta \\-\beta &s-\alpha \end{pmatrix}}{\begin{pmatrix}{\frac {\beta }{s-\alpha }}\\1\end{pmatrix}}} = 1 ( s − α ) 2 + β 2 ( 2 β − β 2 s − α + ( s − α ) ) {\displaystyle ={\frac {1}{(s-\alpha )^{2}+\beta ^{2}}}{\begin{pmatrix}2\beta \\{\frac {-\beta ^{2}}{s-\alpha }}+(s-\alpha )\end{pmatrix}}} = L − 1 [ e α t ] {\displaystyle ={\mathcal {L}}^{-1}[e^{\alpha t}]} ⋅ 1 s 2 + β 2 ( 2 β − β s + s ) {\displaystyle \cdot {\frac {1}{s^{2}+\beta ^{2}}}{\begin{pmatrix}2\beta \\{\frac {-\beta }{s}}+s\end{pmatrix}}} X = L − 1 [ e α t ] ⋅ 2 β s 2 + β 2 {\displaystyle X={\mathcal {L}}^{-1}[e^{\alpha t}]\cdot {\frac {2\beta }{s^{2}+\beta ^{2}}}} この原像は, x = e α t ⋅ 2 sin  β t {\displaystyle x=e^{\alpha t}\cdot 2\sin \beta t} また, Y = L − 1 [ e α t ] 1 s 2 + β 2 ( − β 2 s + s ) {\displaystyle Y={\mathcal {L}}^{-1}[e^{\alpha t}]{\frac {1}{s^{2}+\beta ^{2}}}\left({\frac {-\beta ^{2}}{s}}+s\right)} = L − 1 [ e α t ] 1 s 2 + β 2 s 2 − β 2 s {\displaystyle ={\mathcal {L}}^{-1}[e^{\alpha t}]{\frac {1}{s^{2}+\beta ^{2}}}{\frac {s^{2}-\beta ^{2}}{s}}} これを部分分数展開すると, Y = L − 1 [ e α t ] ( 2 s s 2 + β 2 − 1 s ) {\displaystyle Y={\mathcal {L}}^{-1}[e^{\alpha t}]\left({\frac {2s}{s^{2}+\beta ^{2}}}-{\frac {1}{s}}\right)} この原像は, y = e α t ( 2 cos  β t − 1 ) {\displaystyle y=e^{\alpha t}(2\cos \beta t-1)} ♢ {\displaystyle \diamondsuit }

例107 {\displaystyle \quad }
次の連立微分方程式を解け.
解答例
♢ {\displaystyle \diamondsuit }