45 lines
1.0 KiB
TeX
45 lines
1.0 KiB
TeX
\section*{Problem 1}
|
|
|
|
First, we rearrange the equation.
|
|
|
|
\begin{align*}
|
|
- \frac{d^2u}{dx^2} &= 100 e^{-10x} \\
|
|
\frac{d^2u}{dx^2} &= -100 e^{-10x} \\
|
|
\end{align*}
|
|
|
|
Now we find $u(x)$.
|
|
|
|
% Do the double integral
|
|
\begin{align*}
|
|
u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2 \\
|
|
&= \int \int -100 e^{-10x} dx^2 \\
|
|
&= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
|
|
&= \int 10 e^{-10x} + c_1 dx \\
|
|
&= \frac{10 e^{-10x}}{-10} + c_1 x + c_2 \\
|
|
&= -e^{-10x} + c_1 x + c_2
|
|
\end{align*}
|
|
|
|
Using the boundary conditions, we can find $c_1$ and $c_2$
|
|
|
|
\begin{align*}
|
|
u(0) &= 0 \\
|
|
-e^{-10 \cdot 0} + c_1 \cdot 0 + c_2 &= 0 \\
|
|
-1 + c_2 &= 0 \\
|
|
c_2 &= 1
|
|
\end{align*}
|
|
|
|
\begin{align*}
|
|
u(1) &= 0 \\
|
|
-e^{-10 \cdot 1} + c_1 \cdot 1 + c_2 &= 0 \\
|
|
-e^{-10} + c_1 + c_2 &= 0 \\
|
|
c_1 &= e^{-10} - c_2\\
|
|
c_1 &= e^{-10} - 1\\
|
|
\end{align*}
|
|
|
|
Using the values that we found for $c_1$ and $c_2$, we get
|
|
|
|
\begin{align*}
|
|
u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
|
|
&= 1 - (1 - e^{-10})x - e^{-10x}
|
|
\end{align*}
|