distribusipanas01

\textbf{Penyelesaian :}\\
\begin{enumerate}
\item \textbf{Separasi Variabel}. Jika penyelesaiannya
\begin{equation}
U(x,t) = X(x) T(t)
\end{equation}
maka
\begin{eqnarray*}
U(x,t) &=& X' T \quad U_{xx} = X''T \\
U_t    &=& XT'
\end{eqnarray*}
Persamaan \ref{8.21} menjadi
\begin{equation*}
 XT' = 4 X''T = 2^2 X''T
\end{equation*}
Akibatnya,
\begin{equation*}
\frac{T'}{2^2 T} = \frac{X''}{X} = -\alpha^2
\end{equation*}
\item \textbf{Persamaan Diferensial Biasa}
\begin{eqnarray}
X'' + \alpha^2 X     &=& 0\\
T'  + \alpha^2 2^2 T &=& 0
\end{eqnarray}
\item \textbf{Syarat Homogen}
\begin{equation*}
0 = U(0,t) = X(0) T(t)
\end{equation*}
Karena $T(t)\neq 0$ maka $X(0)=0$
\item \textbf{Persamaan Diferensial} dalam $X$
\begin{equation}
X'' + \alpha^2 X = 0, \quad X(0)=0
\end{equation}
Penyelesaian umum,
\begin{equation*}
X(x) = c_1 \cos \alpha x + c_2 \sin \alpha x
\end{equation*}
Karena $X=0$ diperoleh
\begin{eqnarray*}
X   &=& c_1 \cos (\alpha .0) + c_2 \sin (\alpha. 0) =0 \\
c_1 &=& 0
\end{eqnarray*}
Akibatnya,
\begin{equation*}
X_\alpha(x) = \sin \alpha x , \alpha > 0
\end{equation*}
\item \textbf{Persamaan Diferensial} dalam $T$
\begin{equation}
T' + \alpha^2 . 2^2 T = 0
\end{equation}
Penyelesaiannya,
\begin{equation*}
T_\alpha(t) = e^{-4\alpha^2t}
\end{equation*}
\item \textbf{Himpunan Penyelesaian}
\begin{eqnarray*}
U_{\alpha}(x,t) &=& X_\alpha (x) T_\alpha (t)\\
                &=& e^{-4\alpha^2t} \sin \alpha x ,\alpha >0
\end{eqnarray*}
\item \textbf{Superposisi Integrasi}
\begin{equation}
U(x,t) = \int_0^{\infty} B(\alpha) e^{-4\alpha^2t} \sin \alpha x \text{\, d}\alpha
\end{equation}
dengan
\begin{eqnarray*}
B(\alpha) &=& \frac{2}{\pi} \int_0^{\infty} f(z) \sin \alpha z \text{\, d}z\\
          &=& \frac{2}{\pi} \int_0^{\infty} 1. \sin \alpha z \text{\, d}z\\
          &=& \frac{2}{\pi} \int_0^{\infty} e^{-sz} \sin \alpha z \text{\, d}z ; \quad s=0\\
          &=& \frac{2}{\pi} \left[\mathscr{L}\{\sin \alpha z\}\right] ; \quad s=0\\
          &=& \frac{2}{\pi}. \frac{\alpha}{\alpha^2+s^2}; \quad s=0\\
          &=& \frac{2}{\pi}. \frac{\alpha}{\alpha^2}\\
          &=& \frac{2}{\pi \alpha}
\end{eqnarray*}
Jadi,
\begin{eqnarray*}
U(x,t) &=& \int_0^{\infty} B(\alpha) e^{-4\alpha^2t} \sin \alpha x \text{\, d}\alpha\\
       &=& \frac{2}{\pi} \int_0^{\infty} \frac{\sin \alpha x}{\alpha} e^{-4\alpha^2t} \text{\, d}\alpha
\end{eqnarray*}
\end{enumerate}