The Exact Differential Equation
Chapter Ninety-Two
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 203 to 204 of 303
In one line
An equation is exact when its left-hand side is already the total differential of some function, so the solution is that function equal to a constant.
The idea before the test
Suppose there is a function F(x, y) whose total differential is exactly the left-hand side of the equation.
dF = (dF/dx) dx + (dF/dy) dy
Then the equation M dx plus N dy = 0 says dF = 0, which says F does not change. So F(x, y) = C, and that is the solution.
Nothing was integrated. The answer was already there, written as a differential, and all that was needed was to recognise it.
The simplest possible example
Consider the equation below.
2x dx + 2y dy = 0
The left side is the differential of x squared plus y squared. So the solution is x squared plus y squared = C, a family of circles.
(2x) dx + (2y) dy = 0
x^2 + y^2 = C
You could also have separated the variables and got the same answer. The point is that no integration of a quotient was needed: the left side was a differential already.
What exact means, formally
The equation M(x, y) dx plus N(x, y) dy = 0 is exact if there exists a function F with:
dF/dx = M
dF/dy = N
Then the solution is F = C.
So solving an exact equation is really a matter of recovering F from its two partial derivatives, which is the subject of the chapter after the next.
A worked recognition
Consider the equation below.
(2x + 3y) dx + (3x + 2y) dy = 0
Is there an F whose x-derivative is 2x plus 3y and whose y-derivative is 3x plus 2y?
Try F = x squared plus 3xy plus y squared. Its x-derivative is 2x plus 3y, which matches. Its y-derivative is 3x plus 2y, which matches. So yes.
(2x + 3y) dx + (3x + 2y) dy = 0
x^2 + 3 x y + y^2 = C
Guessing F works for simple equations and is worth trying first. For anything harder there is a procedure, and there is also a test that tells you in advance whether an F exists at all.
Why this is worth a section of the syllabus
Three reasons.
Many equations are exact and it is not obvious. The test takes ten seconds and the solution is then nearly free.
An equation that is not exact can often be made exact, by multiplying through by an integrating factor. That is MU's next label and six chapters of this book, and it turns the exact method into by far the most widely applicable one in section 2.1.
The Exact Differential Equation
Every method in section 2.1 is a special case of it. A separable equation is exact. A linear equation becomes exact after its integrating factor. A homogeneous equation becomes exact after one of the standard factors. So exactness is the idea underneath the whole section rather than one technique among several.
The differential of a product, which you will meet constantly
Recognising a few standard differentials makes exactness visible by eye, and they are worth learning now.
| Expression | Is the differential of |
|---|---|
| x dy + y dx | xy |
| x dy - y dx, over x squared | y/x |
| y dx - x dy, over y squared | x/y |
| 2x dx + 2y dy | x squared plus y squared |
| (x dx + y dy) over (x squared plus y squared) | half the logarithm of (x squared plus y squared) |
| (x dy - y dx) over (x squared plus y squared) | the arctangent of y over x |
The first is the product rule read backwards, and the second and third are the quotient rule. The last two turn up in every homogeneous equation whose answer involves a logarithm or an arctangent, which is why the answers in the previous two chapters looked as they did.
Check yourself
Which of these left-hand sides is a differential, and of what?
| Left side | Differential of |
|---|---|
| y dx + x dy | xy |
| 2xy dx + x squared dy | x squared y |
| cos y dx - x sin y dy | x cos y |
| y dx - x dy | not a differential |
The last row is the one that matters: y dx minus x dy is not exact, and the next chapter's test shows why in one line. It becomes exact after division by y squared, which is the integrating-factor idea arriving early.
(y) dx + (x) dy = 0
x y = C
(2xy) dx + (x^2) dy = 0
x^2 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.