2.6: Exact Equations

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.