User:EGM6321.F11.TEAM1.R.1.7.Playpen

=Problem R.1.7: Proof of linearity=

Find
Show that equation $$(Eq.7.1)$$ is linear

Solution
Proof of Linearity:

In order for equation $$(Eq.7.1)$$ to be linear, the equation:

$$L_2(\alpha u + \beta v) = \alpha L_2(u) + \beta L_2(v)$$  $$\forall u, v,$$   and    $$\forall \alpha, \beta \in \Re$$

 $$(Eq.7.2)$$ must be satisfied. This is known as the superposition principle.

Where $$u$$ and $$v$$ are functions of $$x$$, and $$\alpha$$ and $$\beta$$ are scalars. Next, we define the equation $$L_2(\alpha u + \beta v)$$ as:

$$L_2(\alpha u + \beta v) = \frac{d^2(\alpha u + \beta v)}{dx^2} + a_1 x \frac{d(\alpha u + \beta v)}{dx} + a_o x(\alpha u + \beta v)$$

Grouping $$\alpha $$ and $$\beta $$ terms on both side of the equation yields:

$$\alpha L_2(u) + \beta L_2(v) = \alpha[\frac{d^2 u}{dx^2} + a_1 x\frac{d u}{dx} + a_0 x u] + \beta [\frac{d^2 v}{dx^2} + a_1 x\frac{d v}{dx} + a_0 x v]$$

Substitution of $$u$$ and $$v$$ into $$(Eq.7.1)$$ on the left hand side yields: $$\alpha[\frac{d^2 u}{dx^2} + a_1 x\frac{d u}{dx} + a_0 x u] + \beta [\frac{d^2 v}{dx^2} + a_1 x\frac{d v}{dx} + a_0 x v] = \alpha[\frac{d^2 u}{dx^2} + a_1 x\frac{d u}{dx} + a_0 x u] + \beta [\frac{d^2 v}{dx^2} + a_1 x\frac{d v}{dx} + a_0 x v]$$

Since the superposition principle satisfied, the function is linear.

References:

Problem 3: | Team 4 - Correctly defined differential operator, clear write up.

Problem 4: | Team 6 - Correctly defined differential operator, complete solution.

This solution was created by Ben Neri, Checked by XXXX, and compiled into the complete report by XXXX.