User:Egm6321.f12.team6/report4

=Report 4= ==Problem *4.1: Validation of Exactness Condition for L2-ODE-VC ==

Problem Statement

 * Verify the Exactness of $$ \displaystyle \sqrt{x}y'' + 2xy' + 3y = 0 $$

Given

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

$$ \displaystyle \sqrt{x}y'' + 2xy' + 3y = 0 $$
 * style="width:95%" |
 * style="width:95%" |


 * (6.1)
 * }
 * 1st Exactness condition:
 * {| style="width:100%" border="0"

$$ \displaystyle G(x,y,y',y) = g(x,y,p) + f(x,y,p)y = 0 $$
 * style="width:95%" |
 * style="width:95%" |


 * (6.2)
 * }
 * 2nd Exactness Condition:
 * {| style="width:100%" border="0"

$$ \displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xy} + pg_{yp} - g_{y} $$
 * style="width:95%" |
 * style="width:95%" |


 * (6.3)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp} $$
 * style="width:95%" |
 * style="width:95%" |


 * (6.4)
 * }

Nomenclature

 * L2-ODE - Linear Second Order Differential Equation
 * $$ \displaystyle y' = p $$
 * $$ \displaystyle f_{xx} = \frac {\partial^2 {f}}{\partial{x^2}}; f_{xy} = \frac {\partial^2 {f}}{\partial{x} \partial{y}}; f_{xp} = \frac {\partial^2 {f}}{\partial {x} \partial {p}} = \frac {\partial^2 {f}}{\partial {x} \partial {y'}}$$

Solution

 * Above Equation 6.1 can be expressed as following.
 * $$ \displaystyle \sqrt{x}y'' + 2xp + 3y = 0 $$
 * $$ \displaystyle \underbrace {2xp + 3y}_{\displaystyle g(x,y,p)} + \underbrace {\sqrt{x}}_{\displaystyle f(x,y,p)}y'' = 0 $$
 * Comparing this with Equation 6.2,
 * $$ \displaystyle G(x,y,y',y) = g(x,y,p) + f(x,y,p)y = 0 $$
 * Therefore, this equation passed the first condition of exactness.
 * In order to evaluate 2nd exactness condition, we have to take many different partial differentiation of $$ \displaystyle g(x,y,p) $$, and $$ \displaystyle f(x,y,p) $$.
 * $$ \displaystyle f = \sqrt {x} = x^{\frac {1}{2}} $$
 * $$ \displaystyle f_{x} = \frac{1}{2}x^{-\frac{1}{2}}; f_{y} = 0 $$
 * $$ \displaystyle f_{xx} = -\frac{1}{4}x^{-\frac{3}{2}}; f_{xy} = 0; f_{xp} = 0; f_{yy} = 0; f_{yp} = 0 $$
 * $$ \displaystyle g = 2xp + 3y$$
 * $$ \displaystyle g_{x} = 2p; g_{y} = 3; g_{p} = 2x $$
 * $$ \displaystyle g_{xp} = 2; g_{yp} = 0; g_{pp} = 0$$
 * If we put all corresponding terms into Equation 6.3 and Equation 6.4, then we get two sets of equations.
 * $$ \displaystyle -\frac{1}{4}x^{-\frac{3}{2}} + 0 + 0 \; ? \; 2 + 0 - 3 $$
 * {| style="width:100%" border="0"

$$ \displaystyle -\frac{1}{4}x^{-\frac{3}{2}} \neq -1 $$
 * style="width:95%" |
 * style="width:95%" |


 * (6.5)
 * }
 * $$ \displaystyle 0 + 0 + 0 \; ? \; 0$$
 * {| style="width:100%" border="0"

$$ \displaystyle 0 = 0 $$
 * style="width:95%" |
 * style="width:95%" |


 * (6.6)
 * }
 * Even though Equation 6.4 satisfies as shown from Equation 6.6, Equation 6.3 does not satisfy as shown from Equation 6.5. Therefore, the given equation fails the second condition of exactness.
 * {| style="width:100%" border="0"|-

Since the equation fails to satisfy one of conditions, this equation is not in a form of exact L2-ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Reference
==Problem *4.2:ODEs in Power Form ==

===Problem Statement ===
 * Find $$ \displaystyle m,n \in \mathbb R $$ such that the following N2-ODE is exact:
 * {| style="width:100%" border="0"

$$ \displaystyle \left ( x^{m}y^{n} \right )\left [ \sqrt{x}{y}''+2x{y}'+3y \right ]=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.1)
 * }
 * Show that the first integral is a L1-ODE-VC:
 * {| style="width:100%" border="0"

$$ \displaystyle \phi\left ( x,y,p \right )=xp+\left ( 2x^{\frac {3}{2}}-1 \right )y=k $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.2)
 * }
 * with:
 * {| style="width:100%" border="0"

$$ \displaystyle p(x):=y'(x) $$
 * style="width:95%" |
 * style="width:95%" |


 * }
 * Solve E2.2 for $$ \displaystyle y(x) $$.
 * Solve E2.2 for $$ \displaystyle y(x) $$.

Given

 * All conditions needed is in the problem statement.

Nomenclature

 * L1-ODE-VC is defined as Linear first order ordinary differential equation with varying coefficients.
 * N2-ODE is defined as Nonlinear second order ordinary differential equation.
 * IFM is defined as Euler Integrating factor method.
 * $$ \displaystyle y'(x)=\frac {dy(x)}{dx} $$
 * $$ \displaystyle {y}''(x)=\frac{d^{2}y\left ( x \right )}{dx^{2}} $$
 * $$ \displaystyle f_{xp}=\frac{\partial^{2} f}{\partial x\partial p} $$

Part a):Find $$ \displaystyle m,n $$ that make E2.1 exact

 * N2-ODE is exact under follow two conditions:
 * First condition:
 * E2.1 should fit following form of equation:
 * {| style="width:100%" border="0"

$$ \displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.3)
 * }
 * Second condition:
 * The N2-ODE should satisfy the following equations:
 * {| 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%" |


 * (2.4)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp} $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.5)
 * }
 * With E2.1 we can have:
 * {| style="width:100%" border="0"

$$ \displaystyle \underbrace{2x^{m+1}y^{n}p+3x^{m}y^{n+1}}_{g(x,y,p)}+\underbrace{x^{m+\frac{1}{2}}y^{n}}_{f(x,y,p)}y''=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.6)
 * }
 * Obviously E2.6 fit the form of E2.3 in First condition.
 * With E2.6 we have:
 * {| style="width:100%" border="0"

$$ \displaystyle g(x,y,p)=2x^{m+1}y^{n}p+3x^{m}y^{n+1} $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.7)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle f(x,y,p)=x^{m+\frac{1}{2}}y^{n} $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.8)
 * }
 * With Second condition of exactness, E2.4 and E2.5 should be satisfied.
 * Since E2.8 is not a function of P, we can conclude that $$ \displaystyle f_{xp}=f_{yp}=0 $$.
 * Then E2.5 can be put in following term:
 * {| style="width:100%" border="0"

$$ \displaystyle 2f_{y}=g_{pp} $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.9)
 * }
 * Differentiate E2.8 wrt $$ \displaystyle y $$ and E2.7 wrt $$ \displaystyle p $$:
 * {| style="width:100%" border="0"

$$ \displaystyle f_{y}=nx^{m+\frac{1}{2}}y^{n-1} $$
 * style="width:95%" |
 * style="width:95%" |


 * (2.10)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle g_{pp}=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.11)
 * }
 * Substitute E2.10 and E2.11 into E2.9, we have:
 * {| style="width:100%" border="0"

$$ \displaystyle 2nx^{m+\frac{1}{2}}y^{n-1}=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.12)
 * }
 * Obviously, in E2.12, we have $$ \displaystyle n=0 $$.
 * Differentiate E2.10 and E2.11 we can have following equations:
 * {| style="width:100%" border="0"

$$ \displaystyle f_{xx}=(m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.13)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle f_{xy}=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.14)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle f_{yy}=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.15)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle g_{y}=3x^{m} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.16)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle g_{xp}=2(m+1)x^{m} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.17)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle g_{yp}=0 $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.18)
 * }
 * Substitute E2.13,E2.14,E2.15,E2.16,E2.17 and E2.18 into E2.4, we can have
 * {| style="width:100%" border="0"

$$ \displaystyle (m+\frac{1}{2})(m-\frac{1}{2})x^{(m-\frac{3}{2})}=2(m+1)x^{m}-3x^{m} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.19)
 * }
 * Solving E2.19 we have $$ \displaystyle m=\frac{1}{2} $$.
 * {| style="width:100%" border="0"|-

$$ \displaystyle m=\frac{1}{2}, n=0 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Part b):Show that E2.2 is a L1-ODE-VC

 * With E2.7,E2.8 and the solution of Part a) we have:
 * {| style="width:100%" border="0"

$$ \displaystyle f(x,y,p)=x $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.20)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle g(x,y,p)=2x^{\frac{3}{2}}p+3x^{\frac{1}{2}}y $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.21)
 * }
 * Then we have:
 * {| style="width:100%" border="0"

$$ \displaystyle \phi(x,y,p)=h(x,y)+\int f(x,y,p)dp=h(x,y)+xp+k_{1} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.22)
 * }
 * From E2.20 we have:
 * {| style="width:100%" border="0"

$$ \displaystyle \phi_{x}=h_{x}+p $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.23)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \phi_{y}=h_{y} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.24)
 * }
 * From lecture notes we know that:
 * {| 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%" |


 * <p style="text-align:right">(2.25)
 * }
 * Substitute E2.21,E2.23 and E2.24 into E2.25 we have:
 * {| style="width:100%" border="0"

$$ \displaystyle 2x^{\frac{3}{2}}p+3x^{\frac{1}{2}}y=h_{x}+p+h_{y}p $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.26)
 * }
 * From E2.26 we have:
 * {| style="width:100%" border="0"

$$ \displaystyle h_{x}(x,y)=3x^{\frac{1}{2}}y $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.27)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle h_{y}(x,y)=2x^{\frac{3}{2}}-1 $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.28)
 * }
 * From E2.27 we have:
 * {| style="width:100%" border="0"

$$ \displaystyle h(x,y)=2x^{\frac{3}{2}}y+k_{2}(y) $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.29)
 * }
 * From E2.28 we have:
 * {| style="width:100%" border="0"

$$ \displaystyle h(x,y)=2x^{\frac{3}{2}}y-y+k_{3}(x)$$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.30)
 * }
 * With E2.29 and E2.30 we can conclude that:
 * {| style="width:100%" border="0"

$$ \displaystyle h(x,y)=2x^{\frac{3}{2}}y-y+k_{4} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.31)
 * }
 * Therefore, with E2.22:
 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * {| style="width:100%" border="0"

$$ \displaystyle \phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.32)
 * }
 * }

Part c):Solve E2.2 for $$ \displaystyle y(x) $$

 * From solution Part b), rewrite E2.32:
 * {| style="width:100%" border="0"

$$ \displaystyle y'+(2x^{\frac{1}{2}}-\frac{1}{x})y=\frac{k}{x} $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.33)
 * }
 * Using IFM, we have:
 * {| style="width:100%" border="0"

$$ \displaystyle h_{1}(x)=exp\left [ \int^{x}\left ( 2s^{\frac{1}{2}}-\frac{1}{s} \right )ds+k_{5} \right ]=\frac{1}{x}exp\left[ \frac{4}{3}x^{\frac{3}{2}}+k_{6} \right] $$
 * style="width:95%" |
 * style="width:95%" |


 * <p style="text-align:right">(2.34)
 * }
 * With E2.34 we have:
 * {| style="width:100%" border="0"

$$ \displaystyle y(x)=\frac{1}{h_{1}(x)}\left [ \int^{x}h_{1}(s)(\frac{k}{s})ds+k_{7}\right] $$
 * style="width:95%" |
 * style="width:95%" |

Therefore, we have:
 * <p style="text-align:right">(2.35)
 * }
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=\frac{x}{exp\left[ \frac{4}{3}x^{\frac{3}{2}}+k_{6} \right]}\left [ \int^{x}\frac{1}{s}exp\left[ \frac{4}{3}s^{\frac{3}{2}}+k_{6} \right]\frac{k}{s}ds+k_{7}\right] $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement:

 * Show that the general L2-ODE-VC of the form,


 * Leads to a class of exact L2-ODE-VC, where the first integral is of the form,

Nomenclature

 * $$ p:=y' $$


 * $$ \phi_x:=\frac{\partial \phi}{\partial x} $$


 * $$ \phi_y:=\frac{\partial \phi}{\partial y} $$


 * $$ y':=\frac{dy}{dx} $$


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

Solution
Step 1 - Differentiate, $$ \phi $$, to obtain an explicit form of equation 3.5

Step 2 - Identify the terms of equation 3.3 by comparison to equation 3.6.
 * Because, $$ \phi_x $$, and, $$ \phi_y $$, are both functions of (x, y, p), we cannot positively say;
 * $$ R(x)=\phi_x $$
 * $$ Q(x)=\phi_y $$
 * However, because of the commutative rule of addition we can say;


 * Also, because none are a function of, $$ y'' $$, we can say;


 * Substitute equations 3.7 and 3.8 into equation 3.6 gives us;

Step 3 - By integrating equation 3.8 we can recover, $$ \phi $$.
 * $$ \displaystyle \int \phi_p =\int P(x)dp+k_1(x,y) $$

Step 4 - Solve for, $$ k_1(x,y) $$, from equation 3.10;
 * Differentiate, $$ \phi $$, from equation 3.10 with respect to, $$ x, y $$;


 * From our conclusion reached in equation 3.7, we can substitute equations 3.11, and 3.12 into equation 3.9;


 * By using the same rational to arrive at equation 3.8, we can say; Since, $$ k_1(x,y) $$, is not a function of, $$ p $$, then
 * $$ \frac{dP(x)}{dx}+\frac{\partial k_1(x,y)}{\partial y}=Q(x) $$
 * This in turn allows us to conclude;


 * Integrating equation 3.14 gives us;
 * $$ \int \frac{\partial k_1(x,y)}{\partial x}=\int R(x)ydx+k_2(y) $$

Step 5 - Find, $$ \phi $$
 * Substitute equation 3.15 into equation 3.10


 * We can also define;
 * $$ T(x)=\int R(x)dx $$
 * which leads us to our concluding form of equation 3.2


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


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle \phi(x,y,p)=P(x)p+T(x)y+k $$


 * <p style="text-align:right">(3.17)
 * }

Problem Statement

 * Consider the following L2-ODE-VC
 * $$ G= (cosx)\,y''+(x^2-sinx)\,y'+2xy=0$$
 * 1. Show that it is exact
 * 2. Find $$ \phi $$
 * 3. Solve for $$ y(x) $$

Nomenclature

 * L2-ODE-VC: Linear 2nd Order ODE with varying coefficients
 * $$ \displaystyle p=y'=\frac{dy}{dx}$$
 * $$ \tau, s $$: Dummy variables
 * $$ \displaystyle \phi_x(x,y,p) =\frac{\partial\phi(x,y,p)}{\partial x}, \phi_y(x,y,p) =\frac{\partial \phi(x,y,p)}{\partial y} $$
 * $$ \displaystyle f_{xx}=\frac{\partial^2 f(x,y,p)}{\partial x^2}, f_{xy}=\frac{\partial^2 f(x,y,p)}{\partial x \partial y}, f_{xp}=\frac{\partial^2 f(x,y,p)}{\partial x \partial p} $$
 * $$ \displaystyle f_y=\frac{\partial f(x,y,p)}{\partial y}, f_{yy}=\frac{\partial^2 f(x,y,p)}{\partial y^2}, f_{yp}=\frac{\partial^2 f(x,y,p)}{\partial y \partial p} $$
 * $$ \displaystyle g_y=\frac{\partial g(x,y,p)}{\partial y}, g_{xp}=\frac{\partial^2 g(x,y,p)}{\partial x \partial p}, g_{yp}=\frac{\partial^2 g(x,y,p)}{\partial y \partial p}, g_{pp}=\frac{\partial^2 g(x,y,p)}{\partial p^2} $$

Part 1 Verify exactness

 * Re-write equation (4.1):


 * It follows that


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


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


 * Thus equation(1) satisfies the 1st exactness condition


 * }


 * To verify the second exactness condition, one just need to show that if equation (4.1) satisfies the following equations:


 * Calculate partial derivatives from equation (4.3) and (4.4)
 * $$ f_{xx}=-cosx$$
 * $$ f_{xy}=f_{yy}=g_{yp}=0$$
 * $$ g_{xp}=2x-cosx$$
 * $$ g_{y}=2x$$
 * $$ f_{xp}=f_{yp}=f_{y}=g_{pp}=0$$


 * Bring them into equation (4.5) and (4.6), we have
 * $$ f_{xx}+2p\,f_{xy}+p^2\,f_{yy}=-cosx+0+0=-cosx $$
 * $$ g_{xp}+p\,g_{yp}-g_y=2x-cosx+0-2x=-cosx $$
 * Thus equation (4.5) is satisfied
 * $$ f_{xp}+p\,f_{yp}+2\,f_{y}=0+0+0=0 $$
 * $$ g_{pp}=0 $$
 * Thus equation (4.6) is satisfied


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


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


 * Therefore, equation (4.1) also satisfies the second exactness condition since both equation (4.5) and (4.6) are satisfied, we can conclude that equation (4.1) is an exact L2-ODE-VC.


 * }

Part 2 Find First Integral

 * From equation (4.4), the first integral $$ \phi(x,y,p) $$ can be written as


 * By bringing equation (4.4) into (4.7), we get


 * Further, we can calculate that :$$ \phi_x=h_x-sinx\,p, \phi_y=h_y $$. By bringing them into equation (4.3), we get
 * $$ g(x,y,p)= 2xy+(x^2-sinx)\,p=h_x+(h_y-sinx)\,p$$


 * Integrate both sides of equation (4.9), we get


 * From both equation (4.11) and (4.10), the partial derivative of $$ h(x,y)$$ with respect to $$ y $$ can be expressed as


 * Equation (4.11) can now be written as


 * Bring equation (4.13) into (4.8), we get the first integal:
 * $$ \phi(x,y,p)=x^2\,y+k_1+cosx\,p=k_2 $$


 * <p style="text-align:right">(4.14)
 * }

Part 3 Solve for y(x)

 * The solution in part 2 is indeed a general non-homogeneous L1-ODE-VC


 * Where $$ M(x,y)= x^2\,y, N(x,y)=cosx $$
 * Since $$ M_y= x^2\neq N_x=-sinx $$, we need to find integration factor $$h$$ to make equation (4.15) exact


 * Divide both sides of equation (4.15) by $$ cosx $$, we have


 * Let's assume that $$h$$ is only a function of $$x$$, then


 * Finally, one can derive $$ y(x) $$ with the integrating factor


 * <p style="text-align:right">(4.18)
 * }

Problem Statement

 * Show that these three relations for the symmetry of mixed 2nd partial derivatives are true
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{xy}=\phi_{yx}, \ \ \phi_{yp}=\phi_{py}, \ \ \phi_{px}=\phi_{xp}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.1)
 * }

Given

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

$$\displaystyle g_{o}-\frac{d g_{1}}{dx}+\frac{d^2 g_{2}}{dx^{2}}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.2)
 * }

Nomenclature

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

$$\displaystyle p=y'$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.3)
 * }
 * and
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2} g_{2}}{dx^{2}}=\frac{d}{dx}(\frac{d g_{2}}{dx})$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.4)
 * }

Solution

 * According to the definition,
 * {| style="width:100%" border="0"

$$\displaystyle g_{0}:=\frac{\partial}{\partial y}(\frac{d \phi}{dx})$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.5)
 * }
 * And then
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx} g_{1}=\frac{d}{dx}[\phi_{xy'}+\phi_{yy'}y'+\phi_y+\phi_{y'y'} y'']$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.6)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}}{dx^{2}} g_{2}=\frac{d}{dx}[\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y'']$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.7)
 * }
 * By Substituting Eq. (4.5), (4.6), and (4.7) into Eq. (4.2) we get
 * {| style="width:100%" border="0"

$$\frac{\partial}{\partial y} (\frac{d \phi}{dx})-\frac{d}{dx} [ \phi_{xy'}+\phi_{yy'}y'+\phi_{y}+\phi_{y'y'}y ]+\frac{d}{dx}[ \phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y]=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.8)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\frac{\partial}{\partial y} (\frac{d \phi}{dx})-\frac{d}{dx}(\phi_{xp})+\frac{d}{dx}( \phi_{yp}p)+\frac{d}{dx}(\phi_y)+\frac{d}{dx}(\phi_{pp}y)+\frac{d}{dx}(\phi_{px})+\frac{d}{dx}(\phi_{py}p)+\frac{d}{dx}(\phi_{pp}y)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.9)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle [\frac{\partial}{\partial y}(\frac{d \phi}{dx})-\frac{d}{dx}\phi_{y}]+[\frac{d}{dx}\phi_{px}-\frac{d}{dx}\phi_{xp}]+[\frac{d}{dx}(\phi_{py}p)-\frac{d}{dx}(\phi_{yp}p)]+[\frac{d}{dx}(\phi_{pp}y)-\frac{d}{dx}(\phi_{pp}y)]=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.10)
 * }
 * From Eq.(4.10), we can get
 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx})-\frac{d}{dx}\phi_{y}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.11)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial}{\partial y}(\frac{d \phi}{dx})=\frac{d}{dx}\phi_{y}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.12)
 * }
 * And
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}\phi_{px}-\frac{d}{dx}\phi_{xp}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.13)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}\phi_{px}=\frac{d}{dx}\phi_{xp}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.14)
 * }
 * And
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}(\phi_{py}p)-\frac{d}{dx}(\phi_{yp}p)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.15)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dx}(\phi_{py}p)=\frac{d}{dx}(\phi_{yp}p)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.16)
 * }
 * From Eq.(4.12) ,(4.14) and (4.16), we can get
 * {| style="width:100%" border="0"

$$\displaystyle \phi_{xy}=\phi_{yx}, \ \ \phi_{px}=\phi_{xp}, \ \ \phi_{yp}=\phi_{py}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(4.17)
 * }
 * {| style="width:100%" border="0"|-

It is same as the Eq.(4.1)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Team Contribution

 * {| class="prettytable"


 * colspan="4" | Team Contribution Table
 * colspan="4" | Compiled by: Jiajun Han
 * colspan="4" | Compiled by: Jiajun Han


 * Problems
 * Solved and Typed By
 * Solved and Inspected By
 * Skimmed By
 * Problem 4.1
 * Kyung-Hoon Bang
 * Tianyu Gu
 * All Team Members
 * Problem 4.2
 * Mutian Hu
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 4.3
 * Brandon Taylor
 * Mutian Hu
 * All Team Members
 * Problem 4.4
 * Tianyu Gu
 * Jiajun Han
 * All Team Members
 * Problem 4.5
 * Cheming Lee
 * Brandon Taylor
 * All Team Members
 * }
 * Problem 4.5
 * Cheming Lee
 * Brandon Taylor
 * All Team Members
 * }
 * }