User:Egm6321.f12.team5.yi/report2 2

R2.8 IFM cont'd

Given

Consider the following function
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$$\underbrace{(hM)}_{\displaystyle\color{blue}{\bar M}}+\underbrace{(hN)}_{\displaystyle\color{blue}{\bar N}}\,y'=0$$ (1) Apply 2nd exactness condition to find h:
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$$\bar M_y=\bar N_x$$ (2)
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$$\begin{cases}&\bar M_y=h_y M+h M_y\\&\bar N_x=h_x N+ h N_x\end{cases}$$ (3) Thus for (1) to be exact:
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$$h_x N-h_y M+h(N_x-M_y)=0$$ (4) Solving (4) for the integrating factor h(x,y) is usually not easy.
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Find

Explain why.

Solution

For the equation $$h_x N-h_y M+h(N_x-M_y)=0$$, it contain the partial differential of h both in x an y.