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 }- 2y=4{t}^{2} }$

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

■答

$3$\displaystyle{y=- 2{t}^{2}+2t- 3+\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 2t }+\frac{\hspace{2} 11 \hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ - 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)- s y\left({0}\right) - {y}^{\prime }\left({0}\right)- $ sY\left({s}\right)- y\left({0}\right) $ }$ $3$\displaystyle{ - 2Y\left({s}\right) }$ $3$\displaystyle{ =\frac{\hspace{2}8\hspace{2}}{\hspace{2} {s}^{3} \hspace{2}} }$

初期条件より

$3$\displaystyle{ {s}^{2}Y\left({s}\right)- s+1- $ sY\left({s}\right)- 1 $ }$ $3$\displaystyle{ - 2Y\left({s}\right) }$ $3$\displaystyle{ =\frac{\hspace{2}8\hspace{2}}{\hspace{2} {s}^{3} \hspace{2}} }$

$3$\displaystyle{ $ {s}^{2}- s- 2 $ Y\left({s}\right)=\frac{\hspace{2}8\hspace{2}}{\hspace{2} {s}^{3} \hspace{2}}+s- 2}$

$3$\displaystyle{Y\left({s}\right)=\frac{\hspace{2} \frac{\hspace{2}8\hspace{2}}{\hspace{2} {s}^{3} \hspace{2}}+s- 2 \hspace{2}}{\hspace{2} {s}^{2}- s- 2 \hspace{2}}}$

$3$\displaystyle{Y\left({s}\right)=\frac{\hspace{2} \frac{\hspace{2}8\hspace{2}}{\hspace{2} {s}^{3} \hspace{2}}+s- 2 \hspace{2}}{\hspace{2} $ s- 2 $ $ s+1 $ \hspace{2}}}$

$3$\displaystyle{Y\left({s}\right)=\frac{\hspace{2}8\hspace{2}}{\hspace{2} {s}^{3} $ s- 2 $ $ s+1 $ \hspace{2}}+\frac{\hspace{2} s- 2 \hspace{2}}{\hspace{2} $ s- 2 $ $ s+1 $ \hspace{2}}}$

$3$\displaystyle{Y\left({s}\right)=\frac{\hspace{2}8\hspace{2}}{\hspace{2} {s}^{3} $ s- 2 $ $ s+1 $ \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} s+1 \hspace{2}}}$

ここで，部分分数分解をする．

$3$\displaystyle{ Y\left({s}\right) }$ $3$\displaystyle{ =\frac{\hspace{2} \hspace{1}{k}_{1} \hspace{2}}{\hspace{2}s\hspace{2}}+\frac{\hspace{2} \hspace{1}{k}_{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}+\frac{\hspace{2} \hspace{1}{k}_{3} \hspace{2}}{\hspace{2} {s}^{3} \hspace{2}}+\frac{\hspace{2} \hspace{1}{k}_{4} \hspace{2}}{\hspace{2} s- 2 \hspace{2}}+\frac{\hspace{2} \hspace{1}{k}_{5} \hspace{2}}{\hspace{2} s+1 \hspace{2}} }$ $3$\displaystyle{ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} s+1 \hspace{2}} }$

$3$\displaystyle{=\frac{\hspace{2} \hspace{1}{k}_{1}{s}^{2} $ s- 2 $ $ s+1 $ +\hspace{1}{k}_{2}s $ s- 2 $ $ s+1 $ +\hspace{1}{k}_{3} $ s- 2 $ $ s+1 $ +\hspace{1}{k}_{4}{s}^{3} $ s+1 $ +\hspace{1}{k}_{5}{s}^{3} $ s- 2 $ \hspace{2}}{\hspace{2} {s}^{3} $ s- 2 $ $ s+1 $ \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} s+1 \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} $ \hspace{1}{k}_{1}+\hspace{1}{k}_{4}+\hspace{1}{k}_{5} $ {s}^{4}+ $ - \hspace{1}{k}_{1}+\hspace{1}{k}_{2}+\hspace{1}{k}_{4}- 2\hspace{1}{k}_{5} $ {s}^{3}+ $ - 2\hspace{1}{k}_{1}- \hspace{1}{k}_{2}+\hspace{1}{k}_{3} $ {s}^{2}+ $ - 2\hspace{1}{k}_{2}- \hspace{1}{k}_{3} $ s- 2\hspace{1}{k}_{3} \hspace{2}}{\hspace{2} {s}^{3} $ s- 2 $ $ s+1 $ \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} s+1 \hspace{2}}}$

部分分数分解前と後の分子を係数比較すると

$3$\displaystyle{ \{ \begin{array}{lll} \hspace{1}{k}_{1}+\hspace{1}{k}_{4}+\hspace{1}{k}_{5}=0 & \vspace{6}\\ - \hspace{1}{k}_{1}+\hspace{1}{k}_{2}+\hspace{1}{k}_{4}- 2\hspace{1}{k}_{5}=0 & \vspace{6}\\ - 2\hspace{1}{k}_{1}- \hspace{1}{k}_{2}+\hspace{1}{k}_{3}=0 & \vspace{6}\\ - 2\hspace{1}{k}_{2}- \hspace{1}{k}_{3}=0 & \vspace{6}\\ - 2\hspace{1}{k}_{3}=8 & \vspace{6}\\ \end{array} }$

$3$\displaystyle{ \hspace{1}{k}_{1}=- 3 }$ ， $3$\displaystyle{ \hspace{1}{k}_{2}=2 }$ ， $3$\displaystyle{ \hspace{1}{k}_{3}=- 4 }$ ， $3$\displaystyle{ \hspace{1}{k}_{4}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}} }$ ， $3$\displaystyle{ \hspace{1}{k}_{5}=\frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}} }$

よって

$3$\displaystyle{ Y\left({s}\right)=- \frac{\hspace{2}3\hspace{2}}{\hspace{2}s\hspace{2}}+\frac{\hspace{2}2\hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}- \frac{\hspace{2}4\hspace{2}}{\hspace{2} {s}^{3} \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} s- 2 \hspace{2}} }$ $3$\displaystyle{ +\frac{\hspace{2}8\hspace{2}}{\hspace{2}3\hspace{2}}\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2} s+1 \hspace{2}} }$ $3$\displaystyle{ +\frac{\hspace{2}1\hspace{2}}{\hspace{2} s+1 \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=- 2{t}^{2}+2t- 3+\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 2t }+\frac{\hspace{2} 11 \hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ - t }}$

■別解

微分方程式で解く

特性方程式

$3$\displaystyle{{{\lambda}}^{2}- {\lambda}- 2=0}$

より

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

$3$\displaystyle{{\lambda}=2,- 1}$

よって一般解は

$3$\displaystyle{y=\hspace{1}{C}_{1}{e}^{ 2t }+\hspace{1}{C}_{2}{e}^{ - t }}$ (同次微分方程式の解法を参照)

次に，未定計数法により特殊解を求める．

$3$\displaystyle{ { \left({ \hspace{1}{K}_{2}{t}^{2}+\hspace{1}{K}_{1}t+\hspace{1}{K}_{0} }\right) }^{ \prime \text{}\prime } }$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。
$3$\displaystyle{ - { \left({ \hspace{1}{K}_{2}{t}^{2}+\hspace{1}{K}_{1}t+\hspace{1}{K}_{0} }\right) }^{\prime } }$ $3$\displaystyle{ - 2 $ \hspace{1}{K}_{2}{t}^{2}+\hspace{1}{K}_{1}t+\hspace{1}{K}_{0} $ }$ $3$\displaystyle{ =4{t}^{2} }$

$3$\displaystyle{ - 2\hspace{1}{K}_{2}{t}^{2}- t $ 2\hspace{1}{K}_{2}+2\hspace{1}{K}_{1} $ +2\hspace{1}{K}_{2}- \hspace{1}{K}_{1} }$ $3$\displaystyle{ - 2\hspace{1}{K}_{0} }$ $3$\displaystyle{ =4{t}^{2} }$

$3$\displaystyle{ \{\begin{array}{lll}- 2\hspace{1}{K}_{2}=4& \vspace{6}\\ 2\hspace{1}{K}_{2}+2\hspace{1}{K}_{1}=0& \vspace{6}\\ 2\hspace{1}{K}_{2}-\hspace{1}{K}_{1}- 2\hspace{1}{K}_{0}=0& \vspace{6}\\ \end{array} }$

$3$\displaystyle{\hspace{1}{K}_{2}=- 2,\hspace{1}{K}_{1}=2,\hspace{1}{K}_{0}=- 3}$

よって

$3$\displaystyle{\hspace{1}{y}_{p}=- 2{t}^{2}+2t- 3}$

以上より

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

$3$\displaystyle{{y}^{\prime }=- 4t+2+2\hspace{1}{C}_{1}{e}^{ 2t }- \hspace{1}{C}_{2}{e}^{ - t }}$

初期条件より

$3$\displaystyle{ y\left({0}\right) =- 3+\hspace{1}{C}_{1}+\hspace{1}{C}_{2}}$ ……(1)

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

(1)，(2)より

$3$\displaystyle{\hspace{1}{C}_{1}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}},\hspace{1}{C}_{2}=\frac{\hspace{2}11\hspace{2}}{\hspace{2}3\hspace{2}}}$

よって

$3$\displaystyle{y=- 2{t}^{2}+2t- 3+\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ 2t }+\frac{\hspace{2} 11 \hspace{2}}{\hspace{2}3\hspace{2}}{e}^{ - t }}$

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最終更新日： 2023年6月6日