User:Egm6321.f12.team6.bang/report1

=Report 1=

Problem Statement

 * Show that
 * $$\displaystyle L_2(\cdot)=\frac{d^2(\cdot)}{dx^2}+a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot)$$
 * is Linear

Solution

 * To prove it's linear following equation should be established
 * $$\displaystyle L_2(\alpha u+\beta v)=\alpha L_2(u)+\beta L_2(v)$$
 * Substitue $$\displaystyle (\cdot)$$ with $$\displaystyle (\alpha u+\beta v)$$
 * Then we have
 * $$\displaystyle L_2(\alpha u+\beta v)=\frac{d^2(\alpha u+\beta v)}{dx^2}+a_1(x)\frac{d(\alpha u+\beta v)}{dx}+a_0(x)(\alpha u+\beta v)$$
 * $$=\displaystyle \alpha \frac{d^2(u)}{dx^2}+\beta \frac{d^2(v)}{dx^2}+a_1(x)(\alpha \frac{du}{dx}+\beta \frac{dv}{dx})+a_0(x)(\alpha u+\beta v)$$
 * $$=\displaystyle \alpha \frac{d^2(u)}{dx^2}+a_1(x) \frac{du}{dx}+a_0u(x))+\beta (\frac{d^2(v)}{dx^2}+a_1(x) \frac{dv}{dx}+a_0v(x))$$

Also we have
 * $$\displaystyle L_2(u)=\frac{d^2(u)}{dx^2}+a_1(x) \frac{du}{dx}+a_0u(x)$$
 * $$\displaystyle L_2(v)=\frac{d^2(v)}{dx^2}+a_1(x) \frac{dv}{dx}+a_0v(x)$$
 * Therefore, we can conclude that


 * $$\displaystyle L_2(\alpha u+\beta v)=\alpha L_2(u)+\beta L_2(v)$$
 * is linear