Necesito ayuda con esto. $$Puedes escribirme un c$$digo Python en Jupyter Notebook para graficar una ecuaci$$n de onda, sujeto a las condiciones dadas? Aqu$$ est$$ el ejemplo y su soluci$$n$.$ Haga que el c$$digo sea comprensible y reproducible.$u(0,t)=0,u(1,t)=0u(x,0)=0\mathrm{.}01sen3x,\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{t}}{|}_t=0=06.\{\begin{array}{c}\text{:}[a_0=\frac{1}{}{\displaystyle \underset{-}{\overset{}{}}}f(x)dx=\frac{1}{}{\displaystyle \underset{-}{\overset{0}{}}}{}^{2}dx+\frac{1}{}{\displaystyle \underset{0}{\overset{}{}}}({}^2-x^2)dx=\frac{5}{3}{}^2]\\ a_n=\frac{1}{}{\displaystyle \underset{-}{\overset{}{}}}f(x)cosnxdx=\frac{1}{}{\displaystyle \underset{-}{\overset{0}{}}}{}^{2}cosnxdx+\frac{1}{}{\displaystyle \underset{0}{\overset{}{}}}({}^2-x^2)cosnxdx\\ =\frac{1}{}(\frac{{}^2-x^2}{n}sinnx{|}_0^{}+\frac{2}{n}{\displaystyle \underset{0}{\overset{}{}}}xsinnxdx)=\frac{2}{n^2}(-1{)}^{n+1}\\ b_n=\frac{1}{}{\displaystyle \underset{-}{\overset{}{}}}f(x)sinnxdx=\frac{1}{}{\displaystyle \underset{-}{\overset{0}{}}}{}^{2}sinnxdx+\frac{1}{}{\displaystyle \underset{0}{\overset{}{}}}({}^2-x^2)sinnxdx\\ =\frac{}{n}[(-1{)}^n-1]+\frac{1}{}(\frac{x^2-{}^2}{n}cosnx{|}_0^{}-\frac{2}{n}{\displaystyle \underset{0}{\overset{}{}}}xcosnxdx)=\frac{}{n}(-1{)}^n+\frac{2}{n^3}[1-(-1{)}^n]\\ f(x)=\frac{5{}^2}{6}+{\displaystyle \underset{n=1}{\overset{}{}}}[\frac{2}{n^2}(-1{)}^{n+1}cosnx+(\frac{}{n}(-1{)}^n+\frac{2[1-(-1{)}^n]}{n^3})sinnx]\end{array}$