# 2.6: Exact Equations

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In this section it is convenient to write first order differential equations in the form

$\label{eq:2.5.1} M(x,y)\,dx+N(x,y)\,dy=0.$

This equation can be interpreted as

$\label{eq:2.5.2} M(x,y)+N(x,y)\,{dy\over dx}=0,$

where $$x$$ is the independent variable and $$y$$ is the dependent variable, or as

$\label{eq:2.5.3} M(x,y)\,{dx\over dy}+N(x,y)=0,$

where $$y$$ is the independent variable and $$x$$ is the dependent variable. Since the solutions of Equation \ref{eq:2.5.2} and Equation \ref{eq:2.5.3} will often have to be left in implicit form we will say that $$F(x,y)=c$$ is an implicit solution of Equation \ref{eq:2.5.1} if every differentiable function $$y=y(x)$$ that satisfies $$F(x,y)=c$$ is a solution of Equation \ref{eq:2.5.2} and every differentiable function $$x=x(y)$$ that satisfies $$F(x,y)=c$$ is a solution of Equation \ref{eq:2.5.3}

Here are some examples:

Table $$\PageIndex{1}$$: Examples of Exact Differential Equations in three forms
Equation \ref{eq:2.5.1} Equation \ref{eq:2.5.2} Equation \ref{eq:2.5.3}
$$3x^2y^2\,dx+2x^3y\,dy =0$$ $$3x^2y^2+2x^3y\, {dy\over dx} =0$$ $$3x^2y^2\, {dx\over dy}+2x^3y=0$$
$$(x^2+y^2)\,dx +2xy\,dy=0$$ $$(x^2+y^2)+2xy\, {dy\over dx}=0$$ $$(x^2+y^2)\, {dx\over dy} +2xy=0$$
$$3y\sin x\,dx-2xy\cos x\,dy =0$$ $$3y\sin x-2xy\cos x\, {dy\over dx} =0$$ $$3y\sin x\, {dx\over dy}-2xy\cos x =0$$

Note that a separable equation can be written as Equation \ref{eq:2.5.1} as

$M(x)\,dx+N(y)\,dy=0. \nonumber$

we will develop a method for solving Equation \ref{eq:2.5.1} under appropriate assumptions on $$M$$ and $$N$$. This method is an extension of the method of separation of variables. Before stating it we consider an example.

##### Example $$\PageIndex{1}$$

Show that

$\label{eq:2.5.4} x^4y^3+x^2y^5+2xy=c$

is an implicit solution of

$\label{eq:2.5.5} (4x^3y^3+2xy^5+2y)\,dx+(3x^4y^2+5x^2y^4+2x)\,dy=0.$

###### Solution

Regarding $$y$$ as a function of $$x$$ and differentiating Equation \ref{eq:2.5.4} implicitly with respect to $$x$$ yields

$(4x^3y^3+2xy^5+2y)+(3x^4y^2+5x^2y^4+2x)\,{dy\over dx}=0. \nonumber$

Similarly, regarding $$x$$ as a function of $$y$$ and differentiating Equation \ref{eq:2.5.4} implicitly with respect to $$y$$ yields

$(4x^3y^3+2xy^5+2y){dx\over dy}+(3x^4y^2+5x^2y^4+2x)=0. \nonumber$

Therefore Equation \ref{eq:2.5.4} is an implicit solution of Equation \ref{eq:2.5.5} in either of its two possible interpretations.

You may think Example $$\PageIndex{1}$$ is pointless, since concocting a differential equation that has a given implicit solution is not particularly interesting. However, it illustrates the next important theorem, which we will prove by using implicit differentiation, as in Example $$\PageIndex{1}$$.

##### Theorem $$\PageIndex{1}$$

If $$F=F(x,y)$$ has continuous partial derivatives $$F_x$$ and $$F_y$$, then

$\label{eq:2.5.6} F(x,y)=c$

(with $$c$$ as a constant) is an implicit solution of the differential equation

$\label{eq:2.5.7} F_x(x,y)\,dx+F_y(x,y)\,dy=0.$

Proof

Regarding $$y$$ as a function of $$x$$ and differentiating Equation \ref{eq:2.5.6} implicitly with respect to $$x$$ yields

$F_x(x,y)+F_y(x,y)\,{dy\over dx}=0. \nonumber$

On the other hand, regarding $$x$$ as a function of $$y$$ and differentiating Equation \ref{eq:2.5.6} implicitly with respect to $$y$$ yields

$F_x(x,y)\,{dx\over dy}+F_y(x,y)=0. \nonumber$

Thus, Equation \ref{eq:2.5.6} is an implicit solution of Equation \ref{eq:2.5.7} in either of its two possible interpretations.

We will say that the equation

$\label{eq:2.5.8} M(x,y)\,dx+N(x,y)\,dy=0$

is exact on an an open rectangle $$R$$ if there’s a function $$F=F(x,y)$$ such $$F_x$$ and $$F_y$$ are continuous, and

$\label{eq:2.5.9} F_x(x,y)=M(x,y) \quad \text{and} \quad F_y(x,y)=N(x,y)$

for all $$(x,y)$$ in $$R$$. This usage of “exact” is related to its usage in calculus, where the expression

$F_x(x,y)\,dx+F_y(x,y)\,dy\nonumber$

(obtained by substituting Equation \ref{eq:2.5.9} into the left side of Equation \ref{eq:2.5.8}) is the exact differential of $$F$$.

Example $$\PageIndex{1}$$ shows that it is easy to solve Equation \ref{eq:2.5.8} if it is exact and we know a function $$F$$ that satisfies Equation \ref{eq:2.5.9}. The important questions are:

• Question 1. Given an equation Equation \ref{eq:2.5.8}, how can we determine whether it is exact?
• Question 2. If Equation \ref{eq:2.5.8} is exact, how do we find a function $$F$$ satisfying Equation \ref{eq:2.5.9}?

To discover the answer to Question 1, assume that there’s a function $$F$$ that satisfies Equation \ref{eq:2.5.9} on some open rectangle $$R$$, and in addition that $$F$$ has continuous mixed partial derivatives $$F_{xy}$$ and $$F_{yx}$$. Then a theorem from calculus implies that $\label{eq:2.5.10} F_{xy}=F_{yx}.$ If $$F_x=M$$ and $$F_y=N$$, differentiating the first of these equations with respect to $$y$$ and the second with respect to $$x$$ yields

$\label{eq:2.5.11} F_{xy}=M_y \quad \text{and} \quad F_{yx}=N_x.$

From Equation \ref{eq:2.5.10} and Equation \ref{eq:2.5.11}, we conclude that a necessary condition for exactness is that $$M_y=N_x$$. This motivates the next theorem, which we state without proof.

## Theorem $$\PageIndex{2}$$: The Exactness Condition

Suppose $$M$$ and $$N$$ are continuous and have continuous partial derivatives $$M_y$$ and $$N_x$$ on an open rectangle $$R.$$ Then

$M(x,y)\,dx+N(x,y)\,dy=0\nonumber$

is exact on $$R$$ if and only if

$\label{eq:2.5.12} M_y(x,y)=N_x(x,y)$

for all $$(x,y)$$ in $$R.$$.

To help you remember the exactness condition, observe that the coefficients of $$dx$$ and $$dy$$ are differentiated in Equation \ref{eq:2.5.12} with respect to the “opposite” variables; that is, the coefficient of $$dx$$ is differentiated with respect to $$y$$, while the coefficient of $$dy$$ is differentiated with respect to $$x$$.

##### Example $$\PageIndex{2}$$

Show that the equation

$3x^2y\,dx+4x^3\,dy=0\nonumber$

is not exact on any open rectangle.

###### Solution

Here

$M(x,y)=3x^2y \quad \text{and} \quad N(x,y)=4x^3 \nonumber$

so

$M_y(x,y)=3x^2 \quad \text{and} N_x(x,y)=12 x^2. \nonumber$

Therefore $$M_y=N_x$$ on the line $$x=0$$, but not on any open rectangle, so there’s no function $$F$$ such that $$F_x(x,y)=M(x,y)$$ and $$F_y(x,y)=N(x,y)$$ for all $$(x,y)$$ on any open rectangle.

The next example illustrates two possible methods for finding a function $$F$$ that satisfies the condition $$F_x=M$$ and $$F_y=N$$ if $$M\,dx+N\,dy=0$$ is exact.

##### Example $$\PageIndex{3}$$

Solve

$\label{eq:2.5.13} (4x^3y^3+3x^2)\,dx+(3x^4y^2+6y^2)\,dy=0.$

Solution (Method 1)

Here $M(x,y)=4x^3y^3+3x^2,\quad N(x,y)=3x^4y^2+6y^2,\nonumber$ and $M_y(x,y)=N_x(x,y)=12 x^3y^2\nonumber$ for all $$(x,y)$$. Therefore Theorem $$\PageIndex{2}$$ implies that there’s a function $$F$$ such that

$\label{eq:2.5.14} F_x(x,y)=M(x,y)=4x^3y^3+3x^2$

and

$\label{eq:2.5.15} F_y(x,y)=N(x,y)=3x^4y^2+6y^2$

for all $$(x,y)$$. To find $$F$$, we integrate Equation \ref{eq:2.5.14} with respect to $$x$$ to obtain

$\label{eq:2.5.16} F(x,y)=x^4y^3+x^3+\phi(y),$

where $$\phi (y)$$ is the “constant” of integration. (Here $$\phi$$ is “constant” in that it is independent of $$x$$, the variable of integration.) If $$\phi$$ is any differentiable function of $$y$$ then $$F$$ satisfies Equation \ref{eq:2.5.14}. To determine $$\phi$$ so that $$F$$ also satisfies Equation \ref{eq:2.5.15}, assume that $$\phi$$ is differentiable and differentiate $$F$$ with respect to $$y$$. This yields

$F_y(x,y)=3x^4y^2+\phi'(y). \nonumber$

Comparing this with Equation \ref{eq:2.5.15} shows that

$\phi'(y)=6y^2. \nonumber$

We integrate this with respect to $$y$$ and take the constant of integration to be zero because we are interested only in finding some $$F$$ that satisfies Equation \ref{eq:2.5.14} and Equation \ref{eq:2.5.15}. This yields

$\phi (y)=2y^3. \nonumber$

Substituting this into Equation \ref{eq:2.5.16} yields

$\label{eq:2.5.17} F(x,y)=x^4y^3+x^3+2y^3.$

Now Theorem $$\PageIndex{1}$$ implies that $x^4y^3+x^3+2y^3=c\nonumber$ is an implicit solution of Equation \ref{eq:2.5.13}. Solving this for $$y$$ yields the explicit solution

$y=\left(c-x^3\over2+x^4\right)^{1/3}. \nonumber$

Solution (Method 2)

Instead of first integrating Equation \ref{eq:2.5.14} with respect to $$x$$, we could begin by integrating Equation \ref{eq:2.5.15} with respect to $$y$$ to obtain

$\label{eq:2.5.18} F(x,y)=x^4y^3+2y^3+\psi (x),$

where $$\psi$$ is an arbitrary function of $$x$$. To determine $$\psi$$, we assume that $$\psi$$ is differentiable and differentiate $$F$$ with respect to $$x$$, which yields

$F_x(x,y)=4x^3y^3+\psi'(x). \nonumber$

Comparing this with Equation \ref{eq:2.5.14} shows that

$\psi'(x)=3x^2. \nonumber$

Integrating this and again taking the constant of integration to be zero yields

$\psi(x)=x^3. \nonumber$ Figure $$\PageIndex{1}$$: A direction field and integral curves for $$(4x^3y^3+3x^2)\,dx+(3x^4y^2+6y^2)\,dy=0$$

Substituting this into Equation \ref{eq:2.5.18} yields Equation \ref{eq:2.5.17}.

Figure $$\PageIndex{1}$$ shows a direction field and some integral curves of Equation \ref{eq:2.5.13}.

Here’s a summary of the procedure used in Method 1 of this example. You should summarize procedure used in Method 2.

##### HOWTO: Procedure For Solving An Exact Equation (Method 1)
• Step 1. Check that the equation $\label{eq:2.5.19} M(x,y)\,dx+N(x,y)\,dy=0$ satisfies the exactness condition $$M_y=N_x$$. If not, don’t go further with this procedure.
• Step 2. Integrate ${\partial F(x,y)\over\partial x}=M(x,y)\nonumber$ with respect to $$x$$ to obtain $\label{eq:2.5.20} F(x,y)=G(x,y)+\phi(y),$ where $$G$$ is an antiderivative of $$M$$ with respect to $$x$$, and $$\phi$$ is an unknown function of $$y$$.
• Step 3. Differentiate Equation \ref{eq:2.5.20} with respect to $$y$$ to obtain ${\partial F(x,y)\over\partial y}={\partial G(x,y)\over\partial y}+\phi'(y). \nonumber$
• Step 4. Equate the right side of this equation to $$N$$ and solve for $$\phi'$$; thus, ${\partial G(x,y)\over\partial y}+\phi'(y)=N(x,y), \quad \text{so} \quad \phi'(y)=N(x,y)-{\partial G(x,y)\over\partial y}. \nonumber$
• Step 5. Integrate $$\phi'$$ with respect to $$y$$, taking the constant of integration to be zero, and substitute the result in Equation \ref{eq:2.5.20} to obtain $$F(x,y)$$.
• Step 6. Set $$F(x,y)=c$$ to obtain an implicit solution of Equation \ref{eq:2.5.19}. If possible, solve for $$y$$ explicitly as a function of $$x$$.

It’s a common mistake to omit Step 6 in the procedure above. However, it is important to include this step, since F isn’t itself a solution of Equation \ref{eq:2.5.19}. Many equations can be conveniently solved by either of the two methods used in Example $$\PageIndex{3}$$. However, sometimes the integration required in one approach is more difficult than in the other. In such cases we choose the approach that requires the easier integration.

##### Example $$\PageIndex{4}$$

Solve the equation

$\label{eq:2.5.21} \left( y e ^ { x y } \tan x + e ^ { x y } \sec ^ { 2 } x \right) d x + x e ^ { x y } \tan x \, dy = 0$

###### Solution

We leave it to you to check that $$M_y = N_x$$ on any open rectangle where $$\tan x$$ and $$\sec x$$ are defined. Here we must find a function F such that

$\label{eq:2.5.22} F_x(x, y) = ye^{xy} \tan x + e^{xy} \sec^2 x$

and

$\label{eq:2.5.23} F_y(x, y) = xe^{xy} \tan x.$

It’s difficult to integrate Equation \ref{eq:2.5.22} with respect to $$x$$, but easy to integrate Equation \ref{eq:2.5.23} with respect to $$y$$. This yields

$\label{eq:2.5.24} F(x, y) = e^{xy} \tan x + \psi(x).$

Differentiating this with respect to $$x$$ yields

$F_x(x, y) = y e^{xy} \tan x + e^{xy} \sec^2 x + \psi'(x). \nonumber$

Comparing this with Equation \ref{eq:2.5.22} shows that $$\psi'(x) = 0$$. Hence, $$\psi$$ is a constant, which we can take to be zero in Equation \ref{eq:2.5.24}, and

$e^{xy} \tan x = c, \nonumber$

is an implicit solution of Equation \ref{eq:2.5.21}.

Attempting to apply our procedure to an differential equation that is not exact will lead to failure in Step 4, since the function

$N - \frac { \partial G } { \partial y } \nonumber$

will not be independent of $$x$$ if $$M_y \neq N_x$$, and therefore cannot be the derivative of a function of $$y$$ alone. Example $$\PageIndex{5}$$ illustrates this.

##### Example $$\PageIndex{5}$$

Verify that the equation

$\label{eq:2.5.25} 3x^2y^2\,dx+6x^3y\,dy=0$

is not exact, and show that the procedure for solving exact equations fails when applied to Equation \ref{eq:2.5.25}.

###### Solution

Here $M_y(x,y)=6x^2y \quad \text{and} \quad N_x(x,y)=18x^2y,\nonumber$

so Equation \ref{eq:2.5.25} is not exact. Nevertheless, let us try to find a function $$F$$ such that

$\label{eq:2.5.26} F_x(x,y)=3x^2y^2$

and

$\label{eq:2.5.27} F_y(x,y)=6x^3y.$

Integrating Equation \ref{eq:2.5.26} with respect to $$x$$ yields

$F(x,y)=x^3y^2+\phi(y), \nonumber$

and differentiating this with respect to $$y$$ yields

$F_y(x,y)=2x^3y+\phi'(y). \nonumber$

For this equation to be consistent with Equation \ref{eq:2.5.27},

$6x^3y=2x^3y+\phi'(y), \nonumber$

or

$\phi'(y)=4x^3y. \nonumber$

This is a contradiction, since $$\phi'$$ must be independent of $$x$$. Therefore the procedure fails.

This page titled 2.6: Exact Equations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.