User talk:EGM6321.F12.team5.nguyen R3-8

R*3.8 Construction of a class of N1-ODEs of a specific case

Given

Consider the following functions


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$$\underbrace{\bar b(x,y)c(y)}_{\displaystyle \color{blue}{N(x,y)}}\,y' + \underbrace{a(x)\bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0$$ 13-2 (1)
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$$\frac{h_y}{h}=\frac{1}{M}(N_x-M_y)=:m(y)$$ 11-3 (1)
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This problem was restated from sec. 13: and sec. 11:

Find

Construct a class of N1-ODEs, which is the counterpart of (1) p.13-2, and satisfies the condition (1) p.11-3 that an integrating factor $$ h(y) $$ can be found to render it exact.

Solution This solution was solved on our own. For an N1-ODE to be exact, it must satisfy two conditions: 1) The equation can be written in the particular form:


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$$M(x,y) + N(x,y)y'=0$$ 3.8.1
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From inspection of 13-2 (1), the first exactness condition is satisfied where
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$$M(x,y)= a(x) \bar c(x,y)$$ 3.8.2
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$$ N(x,y)= \bar b(x,y) c(y) $$ 3.8.3
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2) The equation should satisfy the 2nd Exactness Condition:
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$$\bar M_y = \bar N_x$$ 3.8.4
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$$\bar M_y = h_y M + h M_y$$ 3.8.5
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$$\bar N_x= h_x N + h N_x$$ 3.8.6
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$$h_x N - h_y M + h(N_x - M_y)=0$$ 3.8.7
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Expanding on the given equation,
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$$\underbrace{\bar b(x,y)c(y)}_{\displaystyle \color{blue}{N(x,y)}}\,y' + \underbrace{a(x)\bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0$$ 13-2 (1)
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$$\bar b(x,y):= \int^x b(s)ds + k_1(y)$$ 13-2 (2)
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$$\bar c(x,y):= \int^x c(s)ds + k_2(x)$$ 13-3 (1)
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where $$ a(x) $$, $$ b(x) $$ , $$ c(y) $$ are arbitrary functions and $$ k_1(y) $$ , $$ k_2(x) $$  are integration coeffients.

We can find that
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$$M_y(x,y)= a(x)c(y)$$ 3.8.8
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$$M(x,y)= \underbrace{\left[\int^x a(s)ds + k_1(y) \right ]}_{\displaystyle\color{blue}{\bar a(x,y)}}c(y)$$ 3.8.9
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$$N_x(x,y)= b(x)c(y)$$ 3.8.10
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$$N(x,y)= \underbrace{\left[\int^x b(s)ds + k_2(y) \right ]}_{\displaystyle\color{blue}{\bar b(x,y)}}c(y)$$ 3.8.11
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Assuming $$k_1$$ and $$k_2$$ are constants, we can construct a counterpart of (1) p13.2 and satisfy the condition (1) p.11-3 that an integrating factor $$h(y)$$ can be found to render it exact


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$$\frac{h_y}{h}=\frac{1}{M}(N_x-M_y)=\underbrace{\frac{1}{\bar a(x)\cancel{c(y)}} \left[a(x)\cancel{c(y)}-a(x)\cancel{c(y)}) \right ]}_{\displaystyle\color{blue}{m(y)}}$$ 3.8.12
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$$h(y)=exp \left [ \int^x m(s)ds+k \right ]$$ 3.8.13
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