User:Egm6321.f12.team6.Hu.M/report5

Problem *5.8:Equivalence of Two Forms of 2nd Exactness Condition for N2-ODEs
===Problem Statement ===
 * Show that the following equation is equivalent to the 2nd exactness condition given in the Lecture Note 22:
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$$ \displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}-g_{px}-pg_{py}+g_y+(f_{xp}+pf_{yp}+2f_{y}-g_{pp})q=0 $$
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 * (8.1)
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Given

 * 2nd Exactness Condition:
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$$ \displaystyle g_0-\frac{dg_1}{dx}+\frac{d^2g_2}{dx^2}=0 $$
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 * (8.2)
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 * In the equation we have:
 * $$ \displaystyle g_0=\frac{\partial G}{\partial y^{(0)}}=f_yy''+g_y=f_yq+g_y $$
 * $$ \displaystyle g_1=\frac{\partial G}{\partial y^{(1)}}=\frac{\partial G}{\partial p}=f_py''+g_p=f_pq+g_p $$
 * $$ \displaystyle g_2=\frac{\partial G}{\partial q}=f $$

Nomenclature

 * $$ \displaystyle p(x):=y'(x) $$
 * $$ \displaystyle q(x):=y''(x) $$
 * N2-ODEs is defined as Non-linear 2nd-Order Differential Equations.

Solution

 * Now we look at E8.2 and analyze three elements of the equation, we can have:
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$$ \displaystyle g_0=f_yq+g_y $$
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 * (8.3)
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 * $$ \displaystyle \frac{dg_1}{dx}=\frac{d(g_p+f_pq)}{dx} $$
 * $$ \displaystyle =\frac{dg_p}{dx}+\frac{\partial g_p}{\partial y}\frac{dy}{dx}+\frac{\partial g_p}{\partial p}\frac{dp}{dx}+\frac{df_p}{dx}q+\frac{\partial f_p}{\partial y}\frac{dy}{dx}q+\frac{\partial f_p}{\partial p}\frac{dp}{dx}q+f_pq' $$
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$$ \displaystyle =g_{px}+g_{py}p+g_{pp}q+f_{px}q+f_{py}pq+f_{pp}q^2+f_pq' $$
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 * (8.4)
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 * $$ \displaystyle \frac{d^2g_2}{dx^2}=\frac{d}{dx}(\frac{df}{dx}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial p}\frac{dp}{dx})=\frac{d}{dx}(f_x+f_yp+f_pq) $$
 * $$ \displaystyle =\frac{df_x}{dx}+\frac{\partial f_x}{\partial y}\frac{dy}{dx}+\frac{\partial f_x}{\partial p}\frac{dp}{dx}+(\frac{df_y}{dx}+\frac{\partial f_y}{\partial y}\frac{dy}{dx}+\frac{\partial f_y}{\partial p}\frac{dp}{dx})p+f_yq+(\frac{df_p}{dx}+\frac{\partial f_p}{\partial y}\frac{dy}{dx}+\frac{\partial f_p}{\partial p}\frac{dp}{dx})q+f_pq' $$
 * $$ \displaystyle =f_{xx}+f_{xy}p+f_{xp}q+(f_{yx}+f_{yy}p+f_{yp}q)p+f_yq+(f_{px}+f_{py}p+f_{pp}q)q+f_pq' $$
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$$ \displaystyle =f_{xx}+2f_{xy}p+2f_{xp}q+f_{yy}p^2+2f_{yp}pq+f_yq+f_{pp}q^2+f_pq' $$
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 * (8.5)
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 * Substitute E8.3, E8.4 and E8.5 into E8.2 we shall have:
 * $$ \displaystyle f_yq+g_y-g_{px}-g_{py}p-g_{pp}q-f_{px}q-f_{py}pq-f_{pp}q^2-f_pq'+f_{xx}+2f_{xy}p+2f_{xp}q+f_{yy}p^2+2f_{yp}pq+f_yq+f_{pp}q^2+f_pq'=0 $$
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$$ \displaystyle \Rightarrow f_{xx}+2pf_{xy}+p^2f_{yy}-g_{px}-pg_{py}+g_{y}+(f_{px}+pf_{py}+2f_y-g_{pp})q=0 $$
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 * (8.6)
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Obviously we have E8.6 the same as E8.1, thus problem solved.
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