M(x,y) dx + N(x,y)dy = 0

Can always be transformed into an exact DE by multiplying it by some suitable factor, call it r(x,y). This makes the DE exact and is called an

*“integrating factor*”.

*Usually*an appropriate r(x,y) can be found on inspection of the DE and visualizing how it might be most directly simplified, say if both sides were multiplied through by some expression. Example: Find an integrating factor for the DE:

xdy + ydx = x

^{2}y

^{2}dx and solve.

We can rewrite as:

xdy = [x

^{2}y

^{2}–y] dx

and with M = x and N = [x

^{2}y

^{2}–y], we easily see it’s

*not exact*.

Leaving the equation as is, one can see (if one is perceptive) that multiplying both sides by r(x,y) = 1/(x

^{2}y

^{2}) , will work wonders.

First, the right hand side simply becomes dx. E.g.

r(x,y)[xdy + ydx] = dx = 1/(x

^{2}y

^{2}) [ xdy + ydx]

Second, the savvy DE whiz kid (or calculus buff) will quickly spot that the more complicated side of the DE is easily reducible

*via the exact differential form (based on a table of exact differentials):*

1/(x

^{2}y

^{2}) [ xdy + ydx] = d(- 1/xy)

Then, on inspection, our DE quickly reduces to: d(-1/xy) = dx

and integration yields: ò d(-1/xy) = ò dx + c

For which we obtain: -1/xy = x + c

Another more “refined” way to work with integrating factors starts with writing the typical first order linear DE as:

**dy/dx + Py = Q**

And the name of the game is to account for P and Q and also find the integrating factor, r.

Thus, one method for solving the DE shown is to find some function, usually r = r(x) such that if the equation is multiplied by r, the left side becomes the

*derivative of the product*ry. That is:

r(dy/dx) + rPy = rQ

and we then make the effort to impose upon r the condition that:

*r(dy/dx) + rPy = d/dx (ry)*

which is not always easy, but often can be if one is clever enough!

Expanding the right side of the previous eqn. via differentials:

d/dx (ry) = (rdy + y dr) / dx

and adding to the left, gives:

r(dy/dx) + rPy + (-r (dy/dx) – y (dr/dx))

Þ dr/dx = rP

Then, if P = P(x) is a known function, we can solve for r:

Viz. dr/r = Pdx and ln r = ò Pdx + ln C

So: r =

__+__C exp(ò P dx) and C can be taken as

__+__C = 1

Then the function: r = exp(ò Pdx) is called

*the integrating factor*

Example: dy/dx + y = exp(x)

P = 1, Q = exp(x), then: r = exp(òdx) = exp(x)

So: exp(x)y = òexp(2x) + C = exp(2x)/ 2 + c

And y = exp(x)/2 + C exp(-x) or:

y = e

^{x}/2 + Ce

^{-x}

Problems for Math Mavens:

(1) Solve: x

^{2}y dy – xy

^{2}dx – x

^{3}y

^{2}dx = 0

(2) Solve using any method for integrating factors:

x (dy/dx) - 3y = x

^{2 }

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