ラプラス変換に関する問題

■問題

次の式をラプラス変換を用いて，一般解を求めなさい．

$3$\displaystyle{{y}^{\prime\prime }+{y}^{\prime }+y=0}$

(初期条件： $3$\displaystyle{ y\left({0}\right)=0 }$ ， $3$\displaystyle{ {y}^{\prime }\left({0}\right)=1 }$ )

■答

$3$\displaystyle{y=\frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t }\sin \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t}$

■ヒント

ラプラス変換　微分則を用いて解く．

■解き方

ラプラス変換すると

$3$\displaystyle{y{\rightarrow }\limits^{L}Y\left({s}\right)}$

$3$\displaystyle{{y}^{\prime }{\rightarrow }\limits^{L}sY\left({s}\right)- y\left({0}\right)}$

$3$\displaystyle{{y}^{\prime\prime }{\rightarrow }\limits^{L}{s}^{2}Y\left({s}\right)- s\cdot y\left({0}\right)- {y}^{\prime }\left({0}\right)}$

となり，これらを式に代入

$3$\displaystyle{ {s}^{2}Y\left({s}\right)- sy\left({0}\right)- {y}^{\prime }\left({0}\right)+sY\left({s}\right)- y\left({0}\right) }$ $3$\displaystyle{ +Y\left({s}\right) }$ $3$\displaystyle{ =0 }$

初期条件より

$3$\displaystyle{{s}^{2}Y\left({s}\right)- 1+ sY\left({s}\right) +Y\left({s}\right)=0}$

$3$\displaystyle{ $ {s}^{2}+s+1 $ \hspace{1}{Y}_{ $s$ }- 1=0}$

$3$\displaystyle{Y\left({s}\right)=\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left({ s+\frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right)\left({ s+\frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right) \hspace{2}}}$

ここで，部分分数分解をする．

$3$\displaystyle{Y\left({s}\right)=\frac{\hspace{2} \hspace{1}{k}_{1} \hspace{2}}{\hspace{2} s+\frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}+\frac{\hspace{2} \hspace{1}{k}_{2} \hspace{2}}{\hspace{2} s+\frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}}$

$3$\displaystyle{\hspace{1}{k}_{1}={ \lim }\limits_{ s\rightarrow - \frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\left({ s+\frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right)\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left({ s+\frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right)\left({ s+\frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right) \hspace{2}}=\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i}$

$3$\displaystyle{\hspace{1}{k}_{2}={ \lim }\limits_{ s\rightarrow - \frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\left({ s+\frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right)\frac{\hspace{2}1\hspace{2}}{\hspace{2} \left({ s+\frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right)\left({ s+\frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} }\right) \hspace{2}}=- \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i}$

よって

$3$\displaystyle{ Y\left({s}\right)=\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} s+\frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$ $3$\displaystyle{ - \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} s+\frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$

逆ラプラス変換をする　⇒ラプラス変換表はこちら

$3$\displaystyle{y=\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i\left({ {e}^{ - \frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}}t }- {e}^{ - \frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}}t } }\right)}$

$3$\displaystyle{y=\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i\left({ {e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t }\cdot {e}^{ - \frac{\hspace{2} \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}}t }- {e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t }\cdot {e}^{ \frac{\hspace{2} \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}}t } }\right)}$

$3$\displaystyle{y=\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i{e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t }\left({ {e}^{ - \frac{\hspace{2} \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}}t }- {e}^{ \frac{\hspace{2} \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}}t } }\right)}$

オイラーの公式より

$3$\displaystyle{ y=\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}i{e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t } \{ \cos \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t- i\sin \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t }$ $3$\displaystyle{ - $ \cos \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t }$ $3$\displaystyle{ +i\sin \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t $ \} }$

■別解

定数係数線形同次微分方程式の解法を用いて解く．

特性方程式

$3$\displaystyle{ {{\lambda}}^{2}+{\lambda}+1=0 }$

より

$3$\displaystyle{ $ {\lambda}+\frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} $ $ {\lambda}+\frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}} $ =0 }$

$3$\displaystyle{{\lambda}=- \frac{\hspace{2} 1+\sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}},- \frac{\hspace{2} 1- \sqrt{3}i \hspace{2}}{\hspace{2}2\hspace{2}}}$

よって一般解は

$3$\displaystyle{y={e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t } $ \hspace{1}{C}_{1}\cos \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t+\hspace{1}{C}_{2}\sin \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t $ }$ 　　　　　(ただし $3$\displaystyle{\hspace{1}{C}_{1},\hspace{1}{C}_{2}}$ は任意定数)

$3$\displaystyle{ {y}^{\prime }=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t } $ \hspace{1}{C}_{1}\cos \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t+\hspace{1}{C}_{2}\sin \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t $ }$ $3$\displaystyle{ +{e}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}t } $ - \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}\hspace{1}{C}_{1}\sin \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t+\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}\hspace{1}{C}_{2}\cos \frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}t $ }$

初期条件より

$3$\displaystyle{y\left({0}\right)=\hspace{1}{C}_{1}}$ 　･･････(1)

$3$\displaystyle{{y}^{\prime }\left({0}\right)=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\hspace{1}{C}_{1}+\frac{\hspace{2} \sqrt{3} \hspace{2}}{\hspace{2}2\hspace{2}}\hspace{1}{C}_{2}}$ 　･･････(2)

(1)，(2)より

$3$\displaystyle{\hspace{1}{C}_{1}=0,\hspace{1}{C}_{2}=\frac{\hspace{2} 2\sqrt{3} \hspace{2}}{\hspace{2}3\hspace{2}}}$

よって

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最終更新日： 2023年6月6日