User:Oh Isaac/R1-5

=Problem 5=

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)
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becomes


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$$\displaystyle y''+\underbrace{\frac{g'(x)}{g(x)} }_{a_1(x)}y'+a_0(x)y=0$$ (5.2)
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Solution
According to the Chain Rule, equation (5.1) 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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$$\displaystyle X''_i(\xi_i)+\frac{g'_i(\xi_i)}{g_i(\xi_i)}X'_i(\xi_i)+f_i(\xi_i)X_i(\xi_i)=0 $$ (5.3)
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Obviously, When $$\displaystyle y=X_i(\xi_i)$$. equation(5.4) and equation(5.2) are the same.