User:Egm6321.f12.team06.gu/report3

==Problem *3.3: Homogeneous and Particular Solutions ==

Problem Statement

 * For L1-ODE-VC
 * $$y'+a_0(x)\,y=b(x)$$
 * Find its homogeneous solution and particular solution respectively.

Nomenclature

 * L1-ODE-VC: First order linear ODE with varying coefficients
 * $$y'=\frac{dy}{dx}$$
 * $$y_H(x)$$: Homogeneous solution
 * $$y_P(x)$$: Particular solution

Step 1 Find the homogeneous solution

 * The homogeneous counterpart of equation (3.1) is equation (3.2), which can also be denoted as
 * $$\frac{dy}{dx}+ a_0(x)\,y=0$$
 * By separating variables, we have
 * $$\frac{dy}{y}=-a_0(x)\,dx$$
 * By integrating both sides of the equation, we have
 * $$\log(y)=-\int^x a_0(s)\,ds+k_1$$, where $$k_1$$ is a constant
 * Thus, we can draw the homogeneous solution as following
 * $$y_H(x)=e^{-\int^x a_0(s)\,ds+k_1}$$
 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$y_H(x)=k_2\,e^{-\int^x a_0(s)\,ds}$$, where $$\displaystyle k_2=\pm\,e^{k_1}$$


 * }

Step 2 Find the particular solution

 * Denote $$u(x)=k_2$$, then $$y $$ can be written as following


 * Differentiate $$y $$, we have

Bring (3.3) and (3.4) into equation (3.1), we have
 * $$u'(x)\,e^{-\int^x a_0(s)\,ds}-u(x)\,a_0(x)\,e^{-\int^x a_0(s)\,ds}+a_0(x)\,u(x)\,e^{-\int^x a_0(s)\,ds}=b(x)$$
 * $$u'(x)\,e^{-\int^x a_0(s)\,ds}=b(x)$$
 * $$u'(x)=b(x)\,e^{\int^x a_0(s)\,ds}$$

Bring (3.5) into (3.3), we have the general solution for (3.1) as
 * $$y=(\int^x {b(s)\,e^{\int^x a_0(s)\,ds}\,ds}+k_3) \,e^{-\int^x a_0(s)\,ds}$$
 * $$y=\underbrace{(\int^x {b(s)\,e^{\int^x a_0(s)\,ds}\,ds})\,e^{-\int^x a_0(s)\,ds}}_{\color{blue}{y_P(x)}}+ \underbrace{k_3\,e^{-\int^x a_0(s)\,ds}}_{\color{blue}{y_H(x)}}$$
 * Note that the second part of this equation is actually the homogeneous solution for (3.1) if we say $$k_3=k_2$$, while the general solution is a combination of particular solution and homogeneous solution, thus one particular solution for (3.1) can be written as


 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$y_P(x)=(\int^x {b(s)\,e^{\int^x a_0(s)\,ds}\,ds})\,e^{-\int^x a_0(s)\,ds}$$


 * }

Problem Statement

 * Construct a general class of N1-ODEs that is a counterpart of
 * $$\bar b(x,y)c(y)\,y'+a(x) \bar c(x,y)=0$$, where
 * $$\bar b(x,y)=\int^x b(s)\,ds+k_1(y)$$
 * $$\bar c(x,y)=\int^y c(s)\,ds+k_2(x)$$
 * and satisfies the condition
 * $$\frac{h_y}{h}=\frac{1}{M}\,(N_x-M_y)=:\color{blue}{m(y)}$$
 * so that an integrating factor $$h(y)$$ can be found to render it exact.

Given

 * Assumption: $$h_x(x,y)=0$$

Nomenclature

 * N1-ODE: First order nonlinear ODE
 * $$ N_x=\frac{\partial N(x,y)}{\partial x}$$
 * $$ M_y=\frac{\partial M(x,y)}{\partial y}$$

Solution

 * Compare this problem with (1) p.13-2 in lecture note section 13 ,we arbitrarily choose three functions a(x),b(y) and c(y), and construct the following equation:
 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |


 * Where $$\bar a(x,y), \bar c(x,y)$$ are expressed as equation (8.2) and (8.3) respectively.


 * }


 * To prove that an integration factor can be found for equation (8.4):
 * $$N_x=\frac{\partial [\bar a(x,y)b(y)]}{\partial x}$$
 * $$N_x=a(x)\,b(y)$$


 * $$M_y=\frac{\partial [a(x) \bar c(x,y)]}{\partial y}$$
 * $$M_y=a(x)\,c(y)$$


 * Thus equation (8.1) becomes
 * $$\frac{1}{M}\,(N_x-M_y)=\frac{1}{a(x) \bar c(x,y)}\,[a(x)b(y)-a(x)c(y)]$$


 * $$\frac{1}{M}\,(N_x-M_y)=\frac{b(y)-c(y)}{\bar c(x,y)}=m(y)$$, where $$\bar c(x,y)$$ is a function of $$y$$ when $$k_2(x)=d1$$


 * Since $$\frac{1}{M}\,(N_x-M_y)$$ is only a function of $$y$$, according to equation (8.1), we can say that $$h$$ is only a function of $$y$$:
 * Integrating both sides of equation (8.1), we have
 * $$ log(h)=\int^y m(s)ds+k$$


 * Finally, we find the integration factor $$ h(y)$$


 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$ h(y)=e^{\int^y m(s)ds+k}$$


 * }