User:Egm6321.f12.team4.mishra/Homework1 R1.5

=Homework 1 Problem R1.5=

Statement
Show that, $$ \frac{1}{g_i(\xi_i)}\frac{d}{d\xi_i}[g_i(\xi_i)\frac{dX_i(\xi_i)}{d\xi_i}] +

f_i(\xi_i)X_i(\xi_i) = 0 $$

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

Solution:
Change the symbols to simplify ,

$$ \xi_i \rightarrow x $$ ; $$ X_i(\xi_i)\rightarrow y(x) $$ ;

$$ g_i(\xi_i)\rightarrow g(x) $$ ; $$ f_i(\xi_i)\rightarrow a_0(x)

$$.

Hence, the equation becomes,

$$\frac{1}{g(x)} \frac{d}{dx}[g(x)\frac{dy(x)}{dx}] +a_0(x)y(x) = 0 $$

Using chain rule to differentiate the first term on the left hand side,

$$ \frac{1}{g(x)}[g(x)\frac{d^2y(x)}{dx^2}+g'(x)\frac{dy(x)}{dx}]+a_0(x)y(x)=0

$$

Multiplying the terms in square bracket by $$ \frac{1}{g(x)} $$ we get,

$$ \frac{g(x)}{g(x)}

\frac{d^2y(x)}{dx^2}+\frac{g'(x)}{g(x)}\frac{dy(x)}{dx}+a_0(x)y(x)=0 $$

Using $$ \frac{g(x)}{g(x)}=1; \frac{d^2y(x)}{dx^2}=y''; \frac{dy(x)}{dx}=y';

y(x)=y $$

It follows that ,

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