Exact differential equation — test and solve (2xy + 3)dx + (x² - 1)dy = 0

Question

Test whether the differential equation $(2xy + 3)dx + (x^2 - 1)dy = 0$ is exact. If so, solve it.

(JEE Main 2022, similar pattern)

Solution — Step by Step

The equation is in the form $M\,dx + N\,dy = 0$ .

$M = 2xy + 3$ and $N = x^2 - 1$ .

A differential equation is exact if $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$ .

\frac{\partial M}{\partial y} = 2x

\frac{\partial N}{\partial x} = 2x

Since $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} = 2x$ , the equation is exact.

We need $F$ such that $\dfrac{\partial F}{\partial x} = M$ and $\dfrac{\partial F}{\partial y} = N$ .

Integrate $M$ with respect to $x$ :

F = \int (2xy + 3)\,dx = x^2 y + 3x + g(y)

where $g(y)$ is an unknown function of $y$ only.

\frac{\partial F}{\partial y} = x^2 + g'(y) = N = x^2 - 1

g'(y) = -1 \implies g(y) = -y

F(x, y) = x^2 y + 3x - y = C

\mathbf{x^2 y + 3x - y = C}

where $C$ is an arbitrary constant.

Why This Works

An exact differential equation $M\,dx + N\,dy = 0$ means there exists a function $F(x,y)$ such that $dF = M\,dx + N\,dy$ . The equation then simply says $dF = 0$ , so $F = \text{constant}$ .

The exactness condition $\partial M/\partial y = \partial N/\partial x$ comes from the equality of mixed partial derivatives: $\frac{\partial^2 F}{\partial y \partial x} = \frac{\partial^2 F}{\partial x \partial y}$ .

The method reconstructs $F$ by integrating one partial derivative and using the other to pin down the arbitrary function.

Alternative Method

You can also integrate $N$ with respect to $y$ first: $F = \int(x^2 - 1)\,dy = x^2 y - y + h(x)$ . Then $\partial F/\partial x = 2xy + h'(x) = 2xy + 3$ , so $h'(x) = 3$ , $h(x) = 3x$ . Same result: $F = x^2 y - y + 3x = C$ .

When the equation is not exact, look for an integrating factor. If $\frac{M_y - N_x}{N}$ is a function of $x$ only, the integrating factor is $e^{\int \frac{M_y - N_x}{N}\,dx}$ . If $\frac{N_x - M_y}{M}$ is a function of $y$ only, the integrating factor is $e^{\int \frac{N_x - M_y}{M}\,dy}$ .

Common Mistake

After integrating $M$ w.r.t. $x$ , students often write $g(y)$ as a constant instead of a function of $y$ . The “constant of integration” when integrating a partial derivative is not a number — it is a function of the other variable. Missing this function $g(y)$ means the solution will not satisfy $\partial F/\partial y = N$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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