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chphiranya Group Title

Does ant one know how to prove the 'Potential function theorem' used in solving exact differential equations ? Theorem :- If, M(x,y) + N(x,y)(dy/dx) =0 , is an exact equation then, there exists a potential function P s.t. ∂P/∂x=M and ∂P/∂y=N

  • one year ago
  • one year ago

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  1. oldrin.bataku Group Title
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    http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives ... which allows us to note \(\frac{\partial M}{\partial y}=\frac{\partial f}{\partial x\partial y}=\frac{\partial f}{\partial y\partial x}=\frac{\partial N}{\partial x}\). Given \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\), we can construct a potential function \(P(x,y)\) s.t. \(dP=M(x,y)\ dx+N(x,y)\ dy\). Let \(\frac{\partial P}{\partial x}=M(x,y)\), we integrate w.r.t. \(x\) to yield \(P(x,y)=\int M(x,y)\ dx+C(y)\). Differentiate w.r.t. \(y\) to come to \(\frac{\partial P}{\partial y}=\frac\partial{\partial y}\int M(x,y)\ dx+\frac{dC}{dy}\); given \(\frac{\partial P}{\partial y}=N(x,y)\), we thus find \(\frac{dC}{dy}=N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\ dx\). Now that we have eliminated mentions of derivatives of \(P\), we may integrate to yield \(C(y)=\int\left[N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\ dx\right]dy\). We may now substitute back into our earlier equation to yield a solution dependent only on \(M(x,y)\), \(N(x,y)\), \(P(x,y)=\int M(x,y)\ dx+\int\left[N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\ dx\right]dy\). http://www2.fiu.edu/~aladrog/ExactDifferentialEqu.pdf

    • one year ago
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