User:Egm6321.f11.team2.Xia/PR1

=Problem 1.5 - Simplification=

Given
The project equation is as follows : $$\displaystyle \frac{1}{g_{i}(\xi_{i})}\frac{d}{d\xi_{i}}\left[g_{i}(\xi_{i})\frac{dX_{i}(\xi_{i})}{d\xi_{i}}\right]+f_{i}(\xi_{i})X_{i}(\xi_{i})=0$$

Find
Show that the Given equation has the following form: $$\displaystyle y''+\frac{g'(x)}{g(x)}y'+a_{0}(x)y=0$$

Solution
In order to simplify the given equation, we shall first change symbols: $$ \displaystyle \xi_{i}=x,\qquad X_{i}(\xi_{i})=y(x);$$ $$ \displaystyle g_{i}(\xi_{i})=g(x),\qquad f_{i}(\xi_{i})=a_{0}(x).$$ Substitute the above expressions into the given equation: $$ \displaystyle \frac{1}{g(x)}\frac{d}{dx}\left[g(x)\frac{dy(x)}{dx}\right]+a_{0}y(x)=0$$ (5.1) $$ \displaystyle \frac{1}{g(x)}\left[\frac{dg(x)}{dx}\frac{dy(x)}{dx}+g(x)\frac{d^{2}y(x)}{dx^{2}}\right]+a_{0}y(x)=0$$ (5.2) $$ \displaystyle y''+\frac{g'(x)}{g(x)}y'+a_{0}(x)y=0$$ where $$ \frac{dg(x)}{dx}=g'(x), \frac{dy(x)}{dx}=y', \frac{d^{2}y(x)}{dx^{2}}=y''.$$ =Problem 1.7- Linear ODE=

Given
An ODE of 2nd order with variable coefficients is read as $$\displaystyle L_2(y)= \frac{d^2(y)}{dx^2}+a_1(x)\frac{d(y)}{dx}+a_0(x)(y) $$ where $$\displaystyle a_0(x), a_1(x)$$ are variable coefficients

Find
show that $$\displaystyle L_2(\cdot)=\frac{d^2(\cdot)}{dx^2}+a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot) $$   is a linear operator.

Definition of Linear
If $$\displaystyle L_{2} $$ is a linear operator, the it must satisfy $$\displaystyle L_{2}(\alpha u+\beta v)=\alpha L_{2}(u)+\beta L_{2}(v)$$ Where $$\displaystyle u,v $$ are the independent variables, and $$\displaystyle \alpha, \beta \in \mathbb{R}.$$

$$\displaystyle L_{2}(\alpha u+\beta v)=\frac{d^{2}(\alpha u+\beta v)}{dx^{2}}+a_{1}(x)\frac{d(\alpha u+\beta v)}{dx}+a_{0}(x)(\alpha u+\beta v)$$

$$\displaystyle L_{2}(\alpha u+\beta v)=[\frac{d^{2}(\alpha u)}{dx^{2}}+a_{1}(x)\frac{d(\alpha u)}{dx}+a_{0}(x)(\alpha u)]+[\frac{d^{2}(\beta v)}{dx^{2}}+a_{1}(x)\frac{d(\beta v)}{dx}+a_{0}(x)(\beta v)]$$

$$\displaystyle L_{2}(\alpha u+\beta v)=\alpha[\frac{d^{2}(u)}{dx^{2}}+a_{1}(x)\frac{d(u)}{dx}+a_{0}(x)(u)]+\beta[\frac{d^{2}(v)}{dx^{2}}+a_{1}(x)\frac{d(v)}{dx}+a_{0}(x)(v)]$$

$$\displaystyle L_{2}(\alpha u+\beta v)=\alpha L_{2}(u)+\beta L_{2}(v) $$

$$\displaystyle \therefore L_{2}(\cdot)$$ is a linear operator.