User:Egm6321.f12.team06.gu/report1

Problem Statement

 * Show that
 * $$\displaystyle\frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)]\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$
 * is equivalent to
 * $$\displaystyle y''+\frac{g'(x)}{g(x)}\ y'+a_0(x)y=0$$

Given

 * $$\xi_i=x$$
 * $$X_i(\xi_i)=y(x)$$
 * $$g_i(\xi_i)=g(x)$$
 * $$f_i(\xi_i)=a_0(x)$$

Solution

 * We are given that
 * $$\displaystyle\frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$
 * Substitue $$\displaystyle \xi_i, X_i(\xi_i), g_i(\xi_i), f_i(\xi_i)$$
 * with $$\displaystyle x, y(x), g(x), a_0(x)$$ respectively
 * Then we have
 * $$\displaystyle\frac{1}{g(x)}\frac{d}{dx}\left[g(x)\frac{dy(x)}{dx}\right]+a_0(x)y(x)$$
 * $$=\displaystyle\frac{1}{g(x)}\left[\frac{dg(x)}{dx}\frac{dy}{dx}\ +g(x)\frac{d^2y}{dx^2}\right]+a_0(x)y $$
 * $$=\displaystyle\frac{g'(x)}{g(x)}\ y'+\cancelto{1}\frac{g(x)}{g(x)}\ y''+a_0(x)y$$
 * {| style="width:25%" border="0"|-

$$=\displaystyle y''+\frac{g'(x)}{g(x)}\ y'+a_0(x)y$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }


 * Therefore, we can conclude that


 * $$\displaystyle\frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}\left[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}\right]+f_i(\xi_i)X_i(\xi_i)=0$$
 * is equivalent to


 * $$\displaystyle y''+\frac{g'(x)}{g(x)}\ y'+a_0(x)y=0$$