1 Exact Differentials and Integrating Factors J. R. Rice, AM 105b, 6 February 2008 Consider a first order ode which is given in the form dydx=± M(x,y) N(x,y) . To solve such an ode is equivalent, after multiplying by dx, and rearranging, to finding functions y=y(x)satisfying the differential form M(x,y)dx+N(x,y)dy=0 As a special case. Let us suppose that the form M(x,y)dx+N(x,y)dyis a perfectdifferential (sometimes called an exactdifferential). That means that a function f(x,y)exists so that M(x,y)dx+N(x,y)dy=df(x,y), where df(x,y)=±f(x,y)±xdx+±f(x,y) ±y dy. Thus, if such an f(x,y)does exists, which means that Mdx+Ndy=dffor arbitrary ratios of dx to dy,then it must be the case that M(x,y)=±f(x,y)±xand N(x,y)=±f(x,y) ±y , implying that ±M(x,y)±y=±2f(x,y)±y±x=±2 f(x,y)±x±y= ±N(x,y) ±x To turn that around, if Mand Nsatisfy ±M(x,y)±y=±N(x,y) ±x , then it can be shown (please show it) that the differential form M(x,y)dx+N(x,y)dyis perfect, i.e., that a function f(x,y) exists (and is unique to within an additive constant) such that M(x,y)dx+N(x,y)dy=df(x,y). In such case it is trivial to solve dydx=±M(x,y) N(x,y) , and the solutions are just f(x,y)=C(an arbitrarily chosen constant). If we want the specific solution satisfying y = bwhen x= a, we just choose the constant C=f(a,b) . Of course, in general if we are given two functions M(x,y) and N(x,y), it must be anticipated that ±M(x,y)±y²±N(x,y) ±x , and hence that M(x,y)dx+N(x,y)dyis nota perfect

2 differential. Thus the above approach fails. But let us be optimists and ask as follows: Noting that the original ode could have equivalently been written as dydx=± ²(x,y)M(x,y) ²(x,y)N(x,y) for any non-vanishing function ±(x,y), is it possible that a particular ±(x,y)could exist such that the differential form ±(x,y)M(x,y)dx+±(x,y)N(x,y)dyis perfect? -- i.e., that such that a function ˆ f(x,y)

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