By Peter V. O'Neil

While you are searching for this ebook, then you definately already recognize what you're looking for. I simply began studying the book...I am now not discovering it to be very interactive yet....maybe i'm going to as i take advantage of it extra.

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**Extra resources for Advanced Engineering Mathematics**

**Sample text**

This gives co(x , y) = xZy3 + 2e 4y + 2x, as we got before . Not every first-order differential equation is exact . For example, conside r Y + y' _ 30 CHAPTER 1 First-Order Differential Equation s If there were a potential function cp, then we would hav e a* - y' ay = 1, Integrate acp/ax = y with respect to x to get cp(x, y) = xy + g(y) . Substitute this into &p/ay = 1 to get c3y (xy + g( y)) = x+g (y) = 1 . But this can hold only if g' (y) = 1 - x, an impossibility if g is to be independent of x .

Now we know cp to within this function g . Use the fact that we know acp/ay to write • a-Y = 3x2 y 2 + 8e 4*' a (x2y3 +2x + g ( y)) = 3x2 y 2 + g ( y ) • = ay This equation holds if g'(y) = 8e 4)', hence we may choose g(y) = 2e4y. This gives the potential function cp (x , y) = x2 y 3 +2x+2e4' . If we had chosen to integrate acp/ay first, we would have gotten cp(x , y) = f (3x2 y 2 + 8e4y ) dy = x 2 y3 +2e4y + h(x) . Here h can be any function of one variable, because no matter how h(x) is chosen , a (x 2 y 3 + 2e 4' + h(x)) = 3x 2 y 2 + 8e4y , ay as required .

16) . 16) . This solution was "lost", or at least not found, in using the integrating factor as a method of solution . However, y = 3 is not a singular solution because we can include it i n the expression y = 3 + Kx by allowing K = 0 . 16) i s y = 3 + Kx, with K any real number . 1 Separable Equations and Integrating Factor s We will point out a connection between separable equations and integrating factors . The separable equation y' = A(x)B(y) is in general not exact . To see this, write it as A(x)B(y)-y'=0, so in the present context we have M(x, y) = A(x)B(y) and N(x, y) = -1 .