Question Number 196785 by Spillover last updated on 31/Aug/23
Answered by aleks041103 last updated on 03/Oct/23
$${Separation}\:{of}\:{variables} \\ $$$$\Psi\left({x},{t}\right)={X}\left({x}\right){T}\left({t}\right) \\ $$$$\Rightarrow{X}''{T}=−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }{XT}'' \\ $$$$\Rightarrow\frac{{X}''}{{X}}=−\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\:\frac{{T}\:''}{{T}}=−{a}^{\mathrm{2}} \\ $$$$\Rightarrow{X}''+{a}^{\mathrm{2}} {X}=\mathrm{0} \\ $$$$\Rightarrow{X}={sin}\left({ax}+{b}\right) \\ $$$${T}\:''\:−\left({ac}\right)^{\mathrm{2}} {T}=\mathrm{0}\Rightarrow{T}={sinh}\left({act}+{d}\right) \\ $$$$\Rightarrow{XT}={sin}\left({ax}+{b}\right){sinh}\left({act}+{d}\right) \\ $$$${X}\left(\mathrm{0}\right)={X}\left({L}\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left({b}\right)={sin}\left({aL}+{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\mathrm{0},\:{aL}=\pi{k}\Rightarrow{a}=\frac{{k}\pi}{{L}},{k}\in\mathbb{Z} \\ $$$$\Rightarrow\Psi\left({x},{t}\right)=\underset{{k}=−\infty} {\overset{\infty} {\sum}}{A}_{{k}} {sin}\left(\frac{\pi{kx}}{{L}}\right){sinh}\left(\frac{\pi{kct}}{{L}}+{B}_{{k}} \right) \\ $$$${where}\:{A}_{{k}} ,\:{B}_{{k}} \:{are}\:{constants}\:{determined}\:{by} \\ $$$${initial}\:{conditions}. \\ $$