User:Egm6321.f12.team5.hou

Given:
$$\begin{align} \frac{1}{g_i( \xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i) \frac{dX_i(\xi_i)}{d\xi_i} \right ]+f_i(\xi_i)X_i(\xi_i)=0\\\end{align}$$

Find:
Show that $$\begin{align} \frac{1}{g_i( \xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i) \frac{dX_i(\xi_i)}{d\xi_i} \right ]+f_i(\xi_i)X_i(\xi_i)=0\\\end{align}$$ becomes

$$\begin{align} y''+\underbrace{\frac{g'(x)}{g(x)}}_{\color{blue}{a_1(x)}}y'+a_0(x)y=0\\\end{align}$$ is a particular case of

$$\begin{align} \displaystyle P(x)y''+Q(x)y'+R(x)y=F(x)\\\end{align}$$, which in turn can be written as

$$\begin{align}\displaystyle y^{(k)}:=\frac{d^ky(x)}{dx^k}\\\end{align}$$

Solution:
$$\begin{align}\frac{1}{g_i( \xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right] +f_i(\xi_i)X_i(\xi_i)\\\end{align}$$

$$\begin{align}=\frac{1}{g_i(\xi_i)}[\frac{dg_i(\xi_i)}{d\xi_i}\frac{dX_i(\xi_i)}{d\xi_i}+g_i(\xi_i)\frac{d^2X_i(\xi_i)}{d\xi^2_i}]+f_i(\xi_i)X_i(\xi_i)\\\end{align}$$

$$\begin{align}=\frac{d^2X_i(\xi_i)}{d\xi^2_i}+\frac{1}{ g_i(\xi_i)}\frac{dg_i(\xi_i)}{d\xi_i} \frac{dX_i(\xi_i)}{d\xi_i}+f_i(\xi_i)X_i(\xi_i)\\\end{align}$$

$$\begin{align}= X''_i(\xi_i)+\frac{g_i'(\xi_i)}{g_i(\xi_i)}X'_i(\xi_i)+f_i(\xi_i)X_i(\xi_i)\\\end{align}$$

By changing the symbols to

$$\begin{align}\xi_i\rightarrow x\\\end{align}$$

$$\begin{align}X_i(\xi_i)\rightarrow y(x)\\\end{align}$$

$$\begin{align}g_i(\xi_i)\rightarrow g(x)\\\end{align}$$

$$\begin{align}f_i(\xi_i)\rightarrow a_0(x)\\\end{align}$$

We can find following simplified form of the separated equations

$$\begin{align}X_i(\xi_i)+\frac{g_i'(\xi_i)}{g_i(\xi_i)}X'_i(\xi_i)+f_i(\xi_i)X_i(\xi_i)=y+\frac{g'(x)}{g(x)}{y'}+a_0(x)y=0\\\end{align}$$

By further changing the symbols to

$$\begin{align}\frac{g'_i(\xi_i)}{g_i(x)}\rightarrow a_1(x)\\\end{align}$$

We get following equation.

$$\begin{align}y+\frac{g'(x)}{g(x)}{y'}+a_0(x)y=y+a_1(x)y'+a_0(x)y=0\\\end{align}$$

If we set a2(x) =1 and F(x) =0, we get the following equation.

$$\begin{align}y+\frac{g'(x)}{g(x)}{y'}+a_0(x)y=a_2(x)y+a_1(x)y'+a_0(x)y=F(x)\\\end{align}$$

Every term in the equation has a same formation as $$\begin{align}a_k(x)y^{(k)}\\\end{align}$$

$$\begin{align}y^{(k)}:=\frac{d^ky(x)}{dx^k}, k=0,1,2\\\end{align}$$

So the equation can be simplified as following

$$\begin{align}\sum a_k(x)y^{(k)}=F(x), k=0,1,2\\\end{align}$$

Conclusion:
$$\begin{align}\frac{1}{g_i( \xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i) \frac{dX_i(\xi_i)}{d\xi_i} \right ]+f_i(\xi_i)X_i(\xi_i)=0\\\end{align}$$ becomes

$$\begin{align} y''+\underbrace{\frac{g'(x)}{g(x)}}_{\color{blue}{a_1(x)}}y'+a_0(x)y=0\\\end{align}$$ is a particular case of

$$\begin{align} \displaystyle P(x)y''+Q(x)y'+R(x)y=F(x)\\\end{align}$$, which in turn can be written as

$$\begin{align}\displaystyle y^{(k)}:=\frac{d^ky(x)}{dx^k}\\\end{align}$$

Author:
Yu Hou