Check ∂M/∂y = ∂N/∂x. Solve Mdx + Ndy = 0 when exact.

Exact Differential Equation — Test and Solution Method

Question

Solve the differential equation:

(2xy + y^2)\,dx + (x^2 + 2xy)\,dy = 0

Also verify whether the equation is exact before solving. (JEE Main 2024 pattern)

Solution — Step by Step

Write the equation in standard form $M\,dx + N\,dy = 0$ :

M = 2xy + y^2, \qquad N = x^2 + 2xy

We treat $M$ as the coefficient of $dx$ and $N$ as the coefficient of $dy$ . This identification is the only thing that matters before testing exactness.

Compute the partial derivatives:

\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy + y^2) = 2x + 2y

\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2 + 2xy) = 2x + 2y

Since $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ , the equation is exact. We can now integrate directly.

We need a function $F(x, y)$ where:

\frac{\partial F}{\partial x} = M = 2xy + y^2

Integrate with respect to $x$ , treating $y$ as constant:

F = \int (2xy + y^2)\,dx = x^2y + xy^2 + g(y)

The $g(y)$ here is not just a constant — it can depend on $y$ . This is the step most students skip, and it costs them marks.

Differentiate $F$ with respect to $y$ :

\frac{\partial F}{\partial y} = x^2 + 2xy + g'(y)

Set this equal to $N = x^2 + 2xy$ :

x^2 + 2xy + g'(y) = x^2 + 2xy

g'(y) = 0 \implies g(y) = C_1 \text{ (a constant)}

The solution to an exact equation is $F(x, y) = C$ :

\boxed{x^2y + xy^2 = C}

This can also be written as $xy(x + y) = C$ .

Why This Works

When we say an equation $M\,dx + N\,dy = 0$ is exact, we mean there exists some function $F(x, y)$ such that $dF = M\,dx + N\,dy$ . So the equation is literally saying $dF = 0$ , which means $F = \text{constant}$ .

The condition $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ is the mathematical test for this. It comes from Clairaut’s theorem — mixed partials of $F$ must be equal: $F_{xy} = F_{yx}$ .

Once you know the equation is exact, the integration procedure is systematic and guaranteed to work. No guesswork, no integrating factors needed.

Alternative Method — Inspection / Grouping

Sometimes you can spot the exact differential directly by grouping terms. Rewrite:

2xy\,dx + 2xy\,dy + y^2\,dx + x^2\,dy = 0

Notice that $x^2\,dy + y^2\,dx$ is not $d(x^2y)$ — but $x^2\,dy + 2xy\,dx = d(x^2y)$ and $y^2\,dx + 2xy\,dy = d(xy^2)$ .

Group more carefully:

\underbrace{(2xy\,dx + x^2\,dy)}_{d(x^2y)} + \underbrace{(y^2\,dx + 2xy\,dy)}_{d(xy^2)} = 0

d(x^2y) + d(xy^2) = 0 \implies d(x^2y + xy^2) = 0

\therefore\; x^2y + xy^2 = C

Same answer, and in JEE this method is often faster once you recognise the standard exact differentials.

Standard exact differentials worth memorising for JEE:

$d(xy) = x\,dy + y\,dx$
$d(x^2y) = 2xy\,dx + x^2\,dy$
$d\!\left(\dfrac{x}{y}\right) = \dfrac{y\,dx - x\,dy}{y^2}$
$d\!\left(\ln\dfrac{x}{y}\right) = \dfrac{y\,dx - x\,dy}{xy}$

Common Mistake

Forgetting that $g(y)$ is a function, not a constant.

When you integrate $M$ with respect to $x$ , the “constant of integration” can depend on $y$ . Students write $g(y) = C$ immediately without checking — then get the wrong $F$ , and the condition $\dfrac{\partial F}{\partial y} = N$ fails.

Always differentiate your $F$ with respect to $y$ , compare with $N$ , and then solve for $g'(y)$ . If $g'(y)$ still contains $x$ , that means you made an arithmetic error somewhere — go back and recheck.

Question

Solution — Step by Step

Why This Works

Alternative Method — Inspection / Grouping

Common Mistake

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