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