Exact Differential Equations
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Direct answer
Exactness is a test before it is a method: write the first-order equation as M dx + N dy = 0 and check whether ∂M/∂y = ∂N/∂x. When the two cross-partials match, the left side is the exact differential dF of some function F, and the solution is simply F(x, y) = C: integrate M with respect to x holding y constant, then add the terms of N that contain no x. When the test fails, an integrating factor rescues the equation — by formula when (∂M/∂y − ∂N/∂x)/N is a function of x alone, giving IF = e^∫f(x)dx, or by inspection when the pieces regroup into d(xy) or d(x^2 + y^2).
What you must remember
- Exactness test: ∂M/∂y = ∂N/∂x; run it before touching any integration.
- Solution recipe: F = ∫M dx (with y constant) + ∫(x-free terms of N) dy = C — the "constant" of the first integration is a function of y, fixed by N.
- IF by x-function: if (M_y − N_x)/N = f(x) only, multiply the equation by e^∫f(x)dx.
- IF by y-function: if (N_x − M_y)/M = g(y) only, multiply by e^∫g(y)dy.
- Homogeneous IF: when M and N are homogeneous of the same degree, IF = 1/(Mx + Ny) works.
- xy-form IF: when M = y·f(xy) and N = x·g(xy), IF = 1/(Mx − Ny) brings exactness.
- Inspection groups: x dy + y dx = d(xy); x dx + y dy = d(x^2 + y^2)/2; (x dy − y dx)/x^2 = d(y/x) — recognising these shortens many solutions.
One exactness check, then solve
Solve (2xy)dx + (x^2 + 2y)dy = 0. Check first: ∂M/∂y = 2x and ∂N/∂x = 2x — exact. Integrate M with respect to x, treating y as a constant: ∫2xy dx = x^2 y. Now differentiate that with respect to y: 2xy... the partial is x^2, which already accounts for the x^2 inside N; the remainder of N is 2y, so add ∫2y dy = y^2. The solution is x^2 y + y^2 = C. Verification is one line of total differentiation: d(x^2 y + y^2) = 2xy dx + x^2 dy + 2y dy, which is exactly the original equation. When the check fails instead, the rescue has a rhythm: compute (M_y − N_x)/N; if it collapses to a function of x alone, multiply through by the exponential integrating factor and re-run the exactness test — the test must now pass, and candidates who skip the re-check occasionally carry an arithmetic error through to the end.
Where marks leak
JEE Main fields direct exact equations and single-formula integrating factors. JEE Advanced mixes: an equation whose IF must be identified from the homogeneous rule or the xy rule, or one best solved by regrouping into d(xy) before any formula is considered. The recurring losses: treating y as a constant in ∫M dx and then writing "+ C" instead of "+ φ(y)" — the y-function is not optional decoration, it is what N will fix; applying the homogeneous IF when M and N have different degrees; and forgetting that after multiplying by any integrating factor the equation has changed, so the solution built from the old M and N is wrong. One quiet convention: the final answer may be left implicit as F(x, y) = C — attempting to force y explicit often ruins a finished solution.
Frequently asked questions
What is the exactness condition for M dx + N dy = 0?
∂M/∂y = ∂N/∂x — equality of the two mixed partials.
How is the solution of an exact equation assembled?
Integrate M in x (y constant), then add the integral in y of N's x-free terms; equate to C.
When is the integrating factor e^∫f(x)dx available?
When (M_y − N_x)/N depends on x alone — f(x) is precisely that quotient.
What integrating factor works for homogeneous M and N of equal degree?
1/(Mx + Ny), after which the equation becomes exact.
What is the differential of xy?
d(xy) = x dy + y dx — the most frequently regrouped pair in inspection solutions.