Test exactness and recover a potential function
For M(x,y) dx + N(x,y) dy = 0, exactness means there is a function F with Fₓ = M and Fᵧ = N. Where the required partial derivatives are continuous on a suitable simply connected region, compare Mᵧ and Nₓ. Integrating M with respect to x produces an unknown function of y, not merely a numerical constant. Differentiate that partial result in y to identify the missing function.
Worked practice example
Solve (2xy + 3) dx + (x² + 4y) dy = 0.
| Step | Working |
|---|---|
| Check exactness | Mᵧ = 2x and Nₓ = 2x, so the equation is exact. |
| Integrate M in x | F = x²y + 3x + g(y). |
| Match Fᵧ with N | Fᵧ = x² + g′(y) = x² + 4y, so g′(y) = 4y and g(y) = 2y². |
| Implicit solution | x²y + 3x + 2y² = C. |
Check your method
Differentiating the implicit solution recovers both original coefficients. If an initial point is given, substitute it into F = C. Exactness is a test on the two coefficients, not a statement that each can be integrated independently and added; doing that would double-count shared terms.
Try it yourself
Solve (2x + y) dx + (x + 2y) dy = 0 through (1,0).
Show the answer and method
Both cross partials equal 1. A potential is F = x² + xy + y². At (1,0), C = 1; the required solution is x² + xy + y² = 1.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.