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 = x2y2 dx and solve.
We can rewrite as:
xdy = [x2y2 –y] dx
and with M = x and N = [x2y2 –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/(x2y2) , will work wonders.
First, the right hand side simply becomes dx. E.g.
r(x,y)[xdy + ydx] = dx = 1/(x2y2) [ 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/(x2y2) [ 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 = ex/2 + Ce-x
Problems for Math Mavens:
(1) Solve: x2y dy – xy2 dx – x3y2dx = 0
(2) Solve using any method for integrating factors:
x (dy/dx) - 3y = x2