User:Oh Isaac

=Problem 5= included in http://en.wikiversity.org/wiki/User:Egm6321.f12.team7/report1

Given
From the lecture note, show that


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$$\displaystyle \frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$ (5.1) becomes
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$$\displaystyle y''+ \begin{matrix} \underbrace{ \frac{g'(x)}{g(x)} } \\ a_1(x) \end{matrix}y'+a_0(x)y=0$$ (5.2)
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Solution
According to the Chain Rule, equation (6.1) yields
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$$\displaystyle \frac{1}{g_i(\xi_i)}\left[g'_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}+g_i(\xi_i)\frac{d^2X_i(\xi_i)}{d\xi_i^2}\right]+f_i(\xi_i)X_i(\xi_i)=0. $$     (5.3)
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After simplification, it becomes


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$$\displaystyle X''_i(\xi_i)+\frac{g'_i(\xi_i)}{g_i(\xi_i)}X'_i(\xi_i)+f_i(\xi_i)X_i(\xi_i)=0 $$ (5.4)
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Obviously, When $$\displaystyle y=X_i(\xi_i)$$. equation(5.4) and equation(5.2) are the same.