Solving an Exact Equation
Chapter Ninety-Four
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 207 to 209 of 303
In one line
Integrate M with respect to x, then add whatever terms of N have no x in them.
The shortcut, stated first
For an exact equation M dx plus N dy = 0, the solution is:
the integral of M with respect to x, treating y as constant, plus the integral of those terms of N that contain no x, with respect to y, all set equal to C.
That is the form to use in an examination. It avoids the double-counting that the long method invites, and it is quick.
Worked with the shortcut
(2x + 3y) dx + (3x + 2y) dy = 0
Integrate M with respect to x: the integral of 2x plus 3y, treating y as a constant, is x squared plus 3xy.
Now look at N, which is 3x plus 2y. The terms with no x in them: just 2y. Integrate that with respect to y: y squared.
Add them: x squared plus 3xy plus y squared.
(2x + 3y) dx + (3x + 2y) dy = 0
x^2 + 3 x y + y^2 = C
The 3x in N was ignored, and correctly: it has already been accounted for by the 3xy that came from integrating M. That is what the shortcut avoids double-counting.
The long method, and why the shortcut follows from it
The long method recovers F from its two partial derivatives, and it is what a question asking you to "show that the equation is exact and solve it" may want in full.
- Integrate M with respect to x, treating y as constant. This gives F up to a function of y alone, because any function of y alone differentiates to zero with respect to x. Call that unknown function g(y).
- Differentiate the result with respect to y.
- Set it equal to N. Everything containing x must cancel, leaving an equation for g prime of y.
- Integrate to find g.
- The solution is F = C.
On the same example. Step one gives x squared plus 3xy plus g(y). Step two gives 3x plus g prime. Step three sets that equal to 3x plus 2y, so g prime is 2y. Step four gives g = y squared. Step five gives the answer above.
Step three is the check built into the method: if the x terms do not cancel, the equation was not exact and you have made an error, either in the test or in step one.
Worked, a harder one
(3x^2 + 4xy) dx + (2x^2 + 2y) dy = 0
Test first: the y-derivative of M is 4x, the x-derivative of N is 4x. Exact.
Shortcut: integrating M with respect to x gives x cubed plus 2x squared y. The terms of N with no x are 2y, integrating to y squared.
Solving an Exact Equation
(3x^2 + 4xy) dx + (2x^2 + 2y) dy = 0
x^3 + 2 x^2 y + y^2 = C
Worked, with a trigonometric function
(sin(y) + y cos(x)) dx + (x cos(y) + sin(x)) dy = 0
Test: the y-derivative of M is cos y plus cos x. The x-derivative of N is cos y plus cos x. Exact.
Shortcut: integrating M with respect to x gives x sin y plus y sin x. Terms of N with no x: none, because both x cos y and sin x contain an x.
(sin(y) + y cos(x)) dx + (x cos(y) + sin(x)) dy = 0
x sin(y) + y sin(x) = C
The answer is the whole of the first integration, and nothing was added. That happens whenever every term of N contains an x, and it is a sign that you have done it right rather than that you have missed something.
Worked, with exponentials
(y e^(x y) + 2x) dx + (x e^(x y) - 2y) dy = 0
Test: the y-derivative of M is e to the xy plus xy e to the xy. The x-derivative of N is the same. Exact.
Shortcut: integrating M with respect to x gives e to the xy plus x squared. Terms of N with no x: minus 2y, integrating to minus y squared.
(y e^(x y) + 2x) dx + (x e^(x y) - 2y) dy = 0
e^(x y) + x^2 - y^2 = C
The mirror version
You may integrate N with respect to y first instead, and then add the terms of M with no y in them, integrated with respect to x. The answer is the same, and it is worth choosing whichever of the two integrals is easier.
On the trigonometric example: integrating N with respect to y gives x sin y plus y sin x, the same expression, and the terms of M with no y in them are none. Same answer, same work.
The check, which is two differentiations
Differentiate your F with respect to x and see whether you get M. Then with respect to y and see whether you get N. If both hold, the answer is right and no further argument is needed.
That is exactly what the checker does with every one of these blocks, and it is what you should do by hand in the examination.
Check yourself
(2x y) dx + (x^2 + 3 y^2) dy = 0
x^2 y + y^3 = C
(y^2 + 2x) dx + (2 x y) dy = 0
x y^2 + x^2 = C
Solving an Exact Equation
(e^y) dx + (x e^y + 2y) dy = 0
x e^y + y^2 = C
(cos(x) cos(y)) dx + (-sin(x) sin(y)) dy = 0
sin(x) cos(y) = C
For the last one, test it first: the y-derivative of cos x cos y is minus cos x sin y, and the x-derivative of minus sin x sin y is minus cos x sin y. Equal, so exact, and the shortcut then gives the answer in one line.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.