Exact Differential Equations

What does "exact" mean?

Write a first-order ODE in differential form: $M(x, y) \, dx + N(x, y) \, dy = 0$

It is exact if its left side is the total differential of some scalar function $F(x, y)$ — that is, there is a "potential" $F$ with $dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy = M \, dx + N \, dy$

When that happens, the ODE reduces to $dF = 0$, so solutions are simply the level curves $F(x, y) = C.$

No integration by separation, no integrating factor — the work is already done, you just have to discover $F$.

The exactness test

If $F$ exists, then $M = F_x$ and $N = F_y$. Equality of mixed partials ($F_{xy} = F_{yx}$) forces $\boxed{\, \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \,}$

This is both necessary and (on a simply connected domain) sufficient. Check it first — if it fails, the equation is not exact and you need an integrating factor.

The given equation

$(2xy + 3) \, dx + (x^{2} + 4y) \, dy = 0$

So $M(x, y) = 2xy + 3$ and $N(x, y) = x^{2} + 4y$.

Step 1 — Test for exactness

$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy + 3) = 2x$ $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^{2} + 4y) = 2x$

Equal. The equation is exact. ✓

Step 2 — Recover $F$ by integrating $M$ with respect to $x$

Since $F_x = M = 2xy + 3$, integrate with respect to $x$ (treating $y$ as a constant): $F(x, y) = \int (2xy + 3) \, dx = x^{2} y + 3x + g(y)$

The "constant" of integration is actually a function $g(y)$ because $y$ was held fixed — any term depending only on $y$ would vanish under $\partial/\partial x$.

Step 3 — Determine $g(y)$ by matching $F_y = N$

Differentiate $F$ with respect to $y$: $F_y = \frac{\partial}{\partial y}\bigl[x^{2} y + 3x + g(y)\bigr] = x^{2} + g'(y)$

This must equal $N = x^{2} + 4y$: $x^{2} + g'(y) = x^{2} + 4y \implies g'(y) = 4y$ $g(y) = 2 y^{2}$

(Constant of integration can be absorbed into the level-curve constant $C$ at the end.)

Step 4 — Assemble $F$ and write the solution

$F(x, y) = x^{2} y + 3x + 2 y^{2}$

Every solution satisfies the implicit equation $\boxed{\, x^{2} y + 3x + 2 y^{2} = C \,}$

This is a one-parameter family of level curves.

Verification

$\frac{\partial F}{\partial x} = 2xy + 3 = M \quad\checkmark$ $\frac{\partial F}{\partial y} = x^{2} + 4y = N \quad\checkmark$

Or, take the total derivative along a solution curve: $dF = (2xy + 3)\, dx + (x^{2} + 4y)\, dy = 0 \quad\checkmark$

An alternative route — integrate $N$ with respect to $y$ instead

Equally valid and sometimes easier when $N$ is simpler than $M$: $F = \int N \, dy = \int (x^{2} + 4y) \, dy = x^{2} y + 2 y^{2} + h(x)$ Match $F_x = M$: $\; 2xy + h'(x) = 2xy + 3 \implies h'(x) = 3 \implies h(x) = 3x$. So $F = x^{2} y + 2 y^{2} + 3x$ — the same answer.

Initial value problem

If $y(1) = 0$, plug in to get $C$: $1^{2} \cdot 0 + 3 \cdot 1 + 2 \cdot 0^{2} = 3 \implies C = 3$ $x^{2} y + 3x + 2 y^{2} = 3$

You usually cannot solve for $y$ explicitly — implicit form is often the final answer for exact equations.

When the test fails — integrating factor for exactness

If $M_y \ne N_x$, look for $\mu(x)$ or $\mu(y)$ making $\mu M \, dx + \mu N \, dy = 0$ exact:

• $\dfrac{M_y - N_x}{N}$ depends on $x$ only $\implies$ $\mu = \mu(x)$, satisfying $\mu' / \mu = (M_y - N_x)/N$.
• $\dfrac{N_x - M_y}{M}$ depends on $y$ only $\implies$ $\mu = \mu(y)$.

Geometric picture

Think of $F(x, y)$ as a height function. The ODE $dF = 0$ says solutions stay at constant height — they are contour lines of the surface. The differential form $M \, dx + N \, dy$ is the gradient's action, and exactness says that form is conservative.

Common mistakes

• Skipping the exactness test. If $M_y \ne N_x$ the potential doesn't exist and every step below is nonsense.
• Treating $g$ as a constant instead of a function of $y$ when integrating $M$ with respect to $x$. The whole method hinges on keeping the "constant" general enough to depend on the other variable.
• Double-counting terms that appear in both integrations. When you integrate $M$ w.r.t. $x$ you've already captured all terms with an $x$ in them; when you then compute $F_y$ and compare with $N$, only the pure-$y$ terms should remain to determine $g'(y)$.
• Forgetting the = C. The answer is an implicit family, not a function.

Try it in the visualization

Draw the level curves $F(x, y) = C$ as contour lines, with the vector field $(M, N)$ shown as arrows. Watch that the arrows stay tangent to the level curves — that's the exactness condition visualized.