The equation M(x,y)dx + N(x,y)dy = 0 is exact, with ∂M/∂y = ∂N/∂x. What form does its general solution take?
Ay = f(x) + C, found by integrating M with respect to x and solving for y
BF(x, y) = C, where F is a potential function satisfying ∂F/∂x = M and ∂F/∂y = N
CThe solution is a parametric curve (x(t), y(t)) traced by the vector field (M, N)
Dx = g(y) + C, found by integrating N with respect to y and solving for x
Exactness means M dx + N dy is the total differential dF of some function F(x,y). So the equation is really just dF = 0 — saying that F is constant along solution curves. The general solution is therefore F(x,y) = C, an implicit equation defining level curves of F. Option A confuses the method of recovery with the form of the solution; options C and D reflect misunderstandings of what the exactness condition means geometrically.
Question 2 Multiple Choice
When solving an exact equation, you integrate M with respect to x to get F(x, y) = ∫M dx + g(y). A student treats g(y) as an arbitrary constant rather than an unknown function. What goes wrong?
ANothing goes wrong — the constant of integration in a multivariable setting is always a pure constant
BThe solution will satisfy ∂F/∂x = M but will generally fail to satisfy ∂F/∂y = N, making F incorrect
CThe student will find too many solutions because g(y) introduces extra degrees of freedom
DThe integration step itself is invalid unless g(y) is confirmed to be constant by the exactness condition
When integrating M with respect to x in a two-variable setting, the 'constant' of integration can depend on y — any function g(y) has zero partial derivative with respect to x and therefore vanishes from ∂F/∂x. Treating g(y) as a pure constant loses this dependence, so the resulting F will satisfy ∂F/∂x = M but will fail ∂F/∂y = N in all but degenerate cases. The entire point of the second step — differentiating F with respect to y and setting it equal to N — is to determine what g(y) must be to satisfy both conditions simultaneously.
Question 3 True / False
The exactness condition ∂M/∂y = ∂N/∂x is both necessary and sufficient for the equation M dx + N dy = 0 to have a potential function, provided the domain is simply connected.
TTrue
FFalse
Answer: True
Necessity follows from the equality of mixed partials: if F exists with ∂F/∂x = M and ∂F/∂y = N, then ∂M/∂y = ∂²F/∂y∂x = ∂²F/∂x∂y = ∂N/∂x (by Clairaut's theorem for smooth F). Sufficiency on a simply connected domain follows from the fact that there are no 'holes' through which a path integral of M dx + N dy could be path-dependent. The simply connected condition is essential — on a domain with holes (like the punctured plane), ∂M/∂y = ∂N/∂x is necessary but not sufficient.
Question 4 True / False
If ∂M/∂y ≠ ∂N/∂x, the equation M dx + N dy = 0 cannot be solved exactly but can generally be made exact by multiplying through by an appropriate integrating factor.
TTrue
FFalse
Answer: False
While an integrating factor μ(x,y) can sometimes restore exactness, finding one in general requires solving its own partial differential equation, which may be intractable. The practical special cases — where (∂M/∂y − ∂N/∂x)/N depends only on x, or (∂N/∂x − ∂M/∂y)/M depends only on y — cover many textbook problems but are not universally applicable. If neither simplification works, the exact equation framework may simply not be the right tool for that ODE.
Question 5 Short Answer
Why must the unknown term g(y) in the step F(x, y) = ∫M dx + g(y) be an arbitrary function of y rather than a simple constant?
Think about your answer, then reveal below.
Model answer: When integrating with respect to x, any function that depends only on y has zero partial derivative with respect to x and therefore vanishes from ∂F/∂x. A pure constant is just the special case where this y-dependent term happens to be constant, but in general the exact equation requires that ∂F/∂y = N, which places a specific constraint on how F must vary with y. The function g(y) is determined (up to a true constant) by differentiating ∫M dx + g(y) with respect to y, setting it equal to N, and integrating the result.
This step is the multivariable analogue of 'constant of integration' in single-variable calculus, but more subtle: instead of a number, the 'constant' with respect to x-integration can be any function of y. Recognizing this is the conceptual core of the method — it is what makes the two-step process work and why it produces a family of solutions F(x,y) = C rather than a single curve.