# Exact Differential Equations

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## EXACT DEs*

In the last section, we discussed how the multiplication of the $$I.F. = {e^{\int {Pdx} }}$$ on both sides of the linear DE

$\frac{{dy}}{{dx}} + P(x)y = Q(x)$

renders this into an exact DE. We now consider the general case of exact DEs. In particular, we want to see what condition must be satisfied in order that the DE

$M(x,\,y)dx + N(x,\,y)dy = 0 \qquad \qquad \ldots (1)$

is exact.

In order for this DE to be exact, it’s LHS must be expressible as the complete differential of some function $$f(x,\,y),$$ i.e.

$M(x,\,y)dx + N(x,\,y)dy = d\left( {f(x,\,y)} \right) \qquad \qquad \ldots (2)$

Now since the function$$f(x,\,y)$$is a function of both x and y, its total differential is a sum of partical differentials with respect to x and y, i.e.,

$df = \frac{{\partial f}}{{\partial x}}dx + \frac{{\partial f}}{{\partial y}}dy \qquad \qquad \ldots (3)$

Comparing (1) and (2), we have

$\frac{{\partial f}}{{\partial x}} = M, \quad \frac{{\partial f}}{{\partial y}} = N \qquad \qquad \ldots (4)$

This gives

$\frac{{{\partial ^2}f}}{{\partial y\partial x}} = \frac{{\partial M}}{{\partial y}}, \quad \frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{{\partial N}}{{\partial x}}$

For continuous $$f(x,\,y),$$

$\frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{{{\partial ^2}f}}{{\partial y\partial x}}$

and thus for the DE in (1) to be exact, we see that the necessary (and in fact sufficient) condition is

$\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}$

* This section contains advanced material and is optional.

If this condition is satisfied, the DE in (1) reduces to

$df(x,\,y) = 0$

which upon integration leads to the required solution:

$f(x,\,y) = C$

As an example, consider the DE

$3x(xy - 2)dx + ({x^3} + 2y)dy = 0 \qquad \ldots (5)$

We have,

\begin{align}&M(x,\,y) = 3x(xy - 2) \quad \Rightarrow \quad \frac{{\partial M}}{{\partial y}} = 3{x^2}\\&N(x,\,y) = {x^3} + 2y \quad \Rightarrow \quad \frac{{\partial N}}{{\partial x}} = 3{x^2}\end{align}

Since \begin{align}\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}},\end{align} the DE is exact and hence we can find a function $$f(x,y)$$ such that the DE is expressible as $$df(x,\,y) = 0.$$ Let us try to explicitly find this function.

From (4), we have

$\frac{{\partial f}}{{\partial x}} = M \quad \Rightarrow \quad \frac{{\partial f}}{{\partial x}} = 3x(xy - 2)$

Integrating with respect to x, while treating y as a constant, we have

$f(x,\,y) = {x^3}y - 3{x^2} + \phi (y) \qquad \qquad \ldots (6)$

The function $$\phi (y)$$ acts as the arbitrary constant of integration, since y is constant for this integration process.

From (4) again we have

$\frac{{\partial f}}{{\partial y}} = N \qquad \Rightarrow \qquad \frac{{\partial f}}{{\partial y}} = {x^3} + 2y$

Evaluating \begin{align}\frac{{\partial f}}{{\partial y}}\end{align} from (6), we have

\begin{align} & \qquad \;\;{x^3} + \phi '(y) = \frac{{\partial f}}{{\partial y}} = {x^3} + 2y\\&\Rightarrow \quad \phi '(y) = 2y\\&\Rightarrow \quad \phi (y) = {y^2} + C \qquad \cdots (7)\end{align}

Finally substituting (7) into (6), we have

$f(x,\,y) = {x^3}y - 3{x^2} + {y^2} + C$

Thus, the solution to the DE in (5) is

\begin{align}&\qquad \;\;f(x,\,y) = {\rm{constant}}\\ &\Rightarrow \quad {x^3}y - 3{x^2} + {y^2} = {\rm{constant}} \qquad \qquad\ldots (8)\end{align}

You are urged to verify that (8) is indeed the required solution by differentiating (8) and observing that (5) is obtained.

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grade 11 | Questions Set 1
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