The Test for Exactness, and Why It Works
Chapter Ninety-Three
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 205 to 206 of 303
In one line
The equation M dx plus N dy = 0 is exact exactly when the y-derivative of M equals the x-derivative of N.
The test
dM/dy = dN/dx
That is the whole test, it takes two differentiations, and it is both necessary and sufficient on any region without holes in it, which every region in this paper is.
Why it works
If the equation is exact then M is the x-derivative of some F and N is its y-derivative. So the y-derivative of M is the mixed second derivative of F, taken x first and then y, and the x-derivative of N is the same mixed derivative taken the other way round.
And for any function with continuous second derivatives, the two mixed partial derivatives are equal. That is Clairaut's theorem on mixed partials, and it is the reason the test looks the way it does.
So exactness forces the test to hold. The converse, that the test holds only for exact equations, needs the region to have no holes in it, and is proved by constructing F explicitly, which is what the next chapter's procedure does.
Worked, one that passes
(2x + 3y) dx + (3x + 2y) dy = 0
M is 2x plus 3y, so the y-derivative of M is 3.
N is 3x plus 2y, so the x-derivative of N is 3.
They are equal, so the equation is exact.
(2x + 3y) dx + (3x + 2y) dy = 0
x^2 + 3 x y + y^2 = C
Worked, one that fails
(y) dx + (-x) dy = 0
M is y, so the y-derivative of M is 1.
N is minus x, so the x-derivative of N is minus 1.
One is not minus one, so the equation is not exact.
That is the y dx minus x dy of the previous chapter, and the test has confirmed it in one line.
But it is nearly exact. Divide through by y squared and the left side becomes the differential of x over y.
(1/y) dx + (-x/y^2) dy = 0
x/y = C
Testing that one: M is one over y, so the y-derivative is minus one over y squared. N is minus x over y squared, so the x-derivative is minus one over y squared. Equal, so it is exact. The division by y squared was an integrating factor, and that is the whole idea of the next six chapters.
Worked, a longer one
(3x^2 + 4xy) dx + (2x^2 + 2y) dy = 0
The y-derivative of M is 4x. The x-derivative of N is 4x. Equal, so exact.
(3x^2 + 4xy) dx + (2x^2 + 2y) dy = 0
x^3 + 2 x^2 y + y^2 = C
The Test for Exactness, and Why It Works
Getting M and N the right way round
The most common error with this test is differentiating the wrong one.
M is the coefficient of dx, and it is differentiated with respect to y. N is the coefficient of dy, and it is differentiated with respect to x.
So each is differentiated with respect to the other variable. A useful way to keep it: the test compares the two ways of arriving at the same mixed second derivative, so it must cross over.
Applying it to the equations of the earlier chapters
| Equation | M | N | dM/dy | dN/dx | Exact |
|---|---|---|---|---|---|
| 2x dx + 2y dy | 2x | 2y | 0 | 0 | yes |
| y dx + x dy | y | x | 1 | 1 | yes |
| y dx - x dy | y | -x | 1 | -1 | no |
| 2xy dx + x^2 dy | 2xy | x^2 | 2x | 2x | yes |
| y^2 dx + 2xy dy | y^2 | 2xy | 2y | 2y | yes |
| y dx + 2x dy | y | 2x | 1 | 2 | no |
The last two rows differ by a factor on one term, and they are on opposite sides of the line. Exactness is a delicate property: almost any change to an exact equation destroys it, which is why the integrating-factor chapters matter so much.
The order to do things in section 2.1
- Is it separable? Separate.
- Is it exact? Solve as exact.
- Is it homogeneous? y = vx.
- Is it linear? The integrating factor of the linear chapter.
- Otherwise, hunt for an integrating factor.
Testing for exactness is second on the list because it costs ten seconds and, when it passes, the rest of the work is short.
Check yourself
| Equation | Exact |
|---|---|
| (x + y) dx + (x + 2y) dy | yes |
| (x + y) dx + (x - y) dy | yes |
| (x + y) dx + (2x + y) dy | no |
| (sin y) dx + (x cos y) dy | yes |
| (y e to the xy) dx + (x e to the xy) dy | yes |
(x + y) dx + (x + 2y) dy = 0
x^2/2 + x y + y^2 = C
(sin(y)) dx + (x cos(y)) dy = 0
x sin(y) = C
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.