chphiranya 2 years ago 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

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1. oldrin.bataku

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