User:Egm6321.f12.team6/report3

=Report 3= ==Problem *3.1: Euler's Integration Factor ==

Problem Statement: Analysis of the Integration Constant $$ k_1 $$

 * Since the General Non-Homogeneous Equation


 * is a Linear First Order Differential Equation, it should only have one integration constant.
 * Show that the integration constant, $$ k_1 $$, in


 * is not necessary.

Solution
Step 1 - We examine equation 1.3, and by Laws of Exponents, we will isolate the constant. The particular Exponent Law we will use states,
 * $$ e^{x+y}=e^xe^y $$
 * Therefore,


 * $$ \displaystyle h(x)=e^{\int^x a_0(s)ds+k_1} $$

Step 2 - We recognize the term, $$ e^{k_1} $$, is itself a constant, and can be denoted by


 * Therefore,

Step 3 - Next, we will restate equation 1.4 in terms of it's upper integrand bound.

Step 4 - We substitute equation 1.7 into equation 1.8

Step 5 - Since, $$ k_3 $$, is a constant it can be brought out in front of the integral giving us

Step 6 - Simplification eliminates the constant, $$ k_3 $$

Step 7 - We conclude
 * {| style="width:100%" border="0"|-

As seen in equation 1.11,
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle y(x)=\frac{1}{e^{\int^x a_0(s)ds}}\int (e^{\int^x a_0(s)ds})b(x)dx+k_2 $$

the integration constant, $$ k_1 $$, is not necessary.
 * (1.12)
 * }

Problem Statement

 * 1. Derive Eq.(12.1),i.e.,the 2nd relation in the 2nd exactness condition, by differentiating the definition of $$ \displaystyle g(x,y,y^{'})$$ in (3)p.16-4 with respect to $$ \displaystyle p:=y^{'}$$ defined in (2)p.7-3.
 * 2. Derive Eq.(12.2),i.e., the 1st relation in the 2nd exactness condition.
 * 3. Verify that Eq.(12.3) satisfies the 2nd exactness condition.

Given

 * 2nd exactness condition for N2-ODEs
 * {| style="width:100%" border="0"

$$\displaystyle f_{xp} + pf_{yp} + 2 f_y = g_{pp}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.1)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx} + 2pf_{xy} + p^2 f_{fyy} = g_{xp} + p g_{yp} - g_y $$
 * style="width:95%" |
 * style="width:95%" |
 * (12.2)
 * }
 * The N2-ODE
 * {| style="width:100%" border="0"

$$\displaystyle x\left( y^{\prime} \right)^2 + yy^{\prime} + \left( xy \right) y^{''} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * (12.3)
 * }

Nomenclature

 * $$ \displaystyle p := y^{'}=\frac{dy}{dx}$$
 * $$ \displaystyle y^{''}=\frac{d^{2}y}{dx^{2}}$$

Solution

 * 1. Derive Eq.(12.1),i.e.,the 2nd relation in the 2nd exactness condition
 * According to (3)p.16-4
 * {| style="width:100%" border="0"

$$\displaystyle g(x,y,p):=\phi_{x}(x,y,p)+\phi_{y}(x,y,p)p$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.4)
 * }
 * and, (4)p.16-4
 * {| style="width:100%" border="0"

$$\displaystyle f(x,y,p):=\phi_{p}(x,y,p)$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.5)
 * }
 * Differentiate the Eq.(12.4) with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{p}=\phi_{xp}+\phi_{yp}p+\phi_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.6)
 * }
 * Differentiate the Eq.(12.6) with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{pp}=\phi_{xpp}+\phi_{ypp}p+\phi_{yp}+\phi_{yp}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.7)
 * }
 * According to Eq.(12.5)
 * {| style="width:100%" border="0"

$$\displaystyle g_{pp}=f_{xpp}+f_{yp}p+2f_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.8)
 * }
 * {| style="width:100%" border="0"|-

This Eq.(12.8) is equal to the Eq.(12.1)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * 2.Derive Eq.(12.2),i.e., the 1st relation in the 2nd exactness condition
 * According to (1)p.17-3, we can get these equations,
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{xy}=\phi_{yx}, \phi_{yp}=\phi_{py}, \phi_{px}=\phi_{xp}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.9)
 * }
 * First, from Eq.(12.9)$$\displaystyle \phi_{px}=\phi_{xp}$$
 * {| style="width:100%" border="0"

$$\displaystyle \left( \phi_{p} \right)_{x} = \left( \phi_{x} \right)_{p}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.10)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle (g - \phi_y y^{\prime})_{p} =(f)_{x}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.11)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle g_{p} -\phi_{yp} y^{\prime} - \phi_y{} = f_x$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.12)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle g_{p}-(\phi_{p})_{y}p-\phi_y = f_x$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.13)
 * }
 * So,
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{y}=g_{p} - f_{y}p - f_{x}$$
 * style="width:95%" |
 * style="width:95%" |
 * (12.14)
 * }
 * Second, $$\displaystyle (\phi_{y})_{p}=(\phi_{p})_{y}$$
 * and, $$\displaystyle g:=\phi_{x}+\phi_{y}p$$
 * {| style="width:100%" border="0"

$$\displaystyle (\frac{g-\phi_{x}}{p})_{p}=f_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.15)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle -\frac{g-\phi_{x}}{p^2} + \frac{g_{p} - f_{x}}{p}=f_y$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.16)
 * }
 * So,
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{x}=g-p(g_{p}-f_{x})+p^{2}f_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.17)
 * }
 * Then, differentiate the Eq.(12.14) with respect to x
 * {| style="width:100%" border="0"

$$\displaystyle (\phi_{y})_{x} = g_{px}-f_{yx}p-f_{xx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.18)
 * }
 * differentiate the Eq.(12.17) with respect to y
 * {| style="width:100%" border="0"

$$\displaystyle (\phi_{x})_{y}=g_{y}-p(g_{py}-f_{xy})+p^{2} f_{yy}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.19)
 * }
 * According to Eq.(12.9)$$\displaystyle \phi_{xy}=\phi_{yx}$$ we can know,
 * {| style="width:100%" border="0"

$$\displaystyle g_{px}-f_{yx}p-f_{xx}=g_{y}-p(g_{py}-f_{xy})+p^{2} f_{yy}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.20)
 * }
 * After rearrange, we can get equation as follow,
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.21)
 * }
 * {| style="width:100%" border="0"|-

This Eq.(12.21) is equal to the Eq.(12.2)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }
 * 3.Verify that Eq.(12.3) satisfies the 2nd exactness condition
 * Suppose
 * {| style="width:100%" border="0"

$$\displaystyle f=xy$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.22)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle g=yy^{'}+x(y^{'})^{2}=yp+xp^{2}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.23)
 * }
 * Differentiate the Eq.(12.23)with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{p}=y+2xp$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.24)
 * }
 * Differentiate the Eq.(12.24)with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{pp}=2x$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.25)
 * }
 * Differentiate the Eq.(12.23)with respect to y:
 * {| style="width:100%" border="0"

$$\displaystyle g_{y}=p$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.26)
 * }
 * Differentiate the Eq.(12.26)with respect to p:
 * {| style="width:100%" border="0"

$$\displaystyle g_{yp}=1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.27)
 * }
 * Differentiate the Eq.(12.22)with respect to x and y:
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx}=f_{yy}=f_{yp}=f_{xp}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.28)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle f_{y}=x $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.29)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle f_{xy}=1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.30)
 * }
 * The 2nd exactness condition satisfies:
 * {| style="width:100%" border="0"

$$\displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=2p$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.31)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle g_{xp}+pg_{yp}-g_{y}=2p$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.32)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle f_{xp}+2f_{y}+f_{yp}p=2x=g_{pp}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(12.33)
 * }

Team Contribution

 * {| class="prettytable"


 * colspan="4" | Team Contribution Table
 * colspan="4" | Compiled by: Tianyu Gu
 * colspan="4" | Compiled by: Tianyu Gu


 * Problems
 * Solved and Typed By
 * Solved and Inspected By
 * Skimmed By
 * Problem 3.1
 * Brandon Taylor
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 3.2
 * Mutian Hu
 * Tianyu Gu
 * All Team Members
 * Problem 3.3
 * Tianyu Gu
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 3.4
 * Brandon Taylor
 * Tianyu Gu
 * All Team Members
 * Problem 3.5
 * Mutian Hu
 * Brandon Taylor
 * All Team Members
 * Problem 3.6
 * Kyung-Hoon Bang
 * Brandon Taylor
 * All Team Members
 * Problem 3.7
 * Kyung-Hoon Bang
 * CheMing Lee
 * All Team Members
 * Problem 3.8
 * Tianyu Gu
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 3.9
 * Jiajun Han
 * Brandon Taylor
 * All Team Members
 * Problem 3.10
 * Kyung-Hoon Bang
 * CheMing Lee
 * All Team Members
 * Problem 3.11
 * CheMing Lee
 * Mutian Hu
 * All Team Members
 * Problem 3.12
 * CheMing Lee
 * Mutian Hu
 * All Team Members
 * }
 * All Team Members
 * Problem 3.11
 * CheMing Lee
 * Mutian Hu
 * All Team Members
 * Problem 3.12
 * CheMing Lee
 * Mutian Hu
 * All Team Members
 * }
 * All Team Members
 * }
 * }