
# 2.7: Exact Differential Equations

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Consider the equation

$f(x,y) = C.$

$f_x(x,y)\hat{\textbf{i}} + f_y(x,y)\hat{\textbf{j}} = 0.$

We can write this equation in differential form as

$f_x(x,y)\, dx+ f_y(x,y)\, dy = 0.$

Now divide by $$dx$$ (we are not pretending to be rigorous here) to get

$f_x(x,y)+ f_y(x,y) \dfrac{dy}{dx} = 0.$

Which is a first order differential equation. The goal of this section is to go backward. That is if a differential equation if of the form above, we seek the original function $$f(x,y)$$ (called a potential function). A differential equation with a potential function is called exact. If you have had vector calculus, this is the same as finding the potential functions and using the fundamental theorem of line integrals.

Example $$\PageIndex{1}$$

Solve

$4xy + 1 + (2x^2 + \cos y)y' = 0.$

Solution

We seek a function $$f(x,y)$$ with

$f_x(x,y) = 4xy + 1$

and

$f_y(x,y) = 2x^2 + \cos y.$

Integrate the first equation with respect to $$x$$ to get

$f(x,y) = 2x^2y + x + C(y) .$

Notice since $$y$$ is treated as a constant, we write $$C(y)$$. Now take the partial derivative with respect to $$y$$ to get

$f_y(x,y) = 2x^2 + C'(y) .$

We have two formulae for $$f_y(x,y)$$ so we can set them equal to each other.

$2x^2 + \cos y = 2x^2 + C'(y)$

That is

$C'(y) = \cos\, y$

or

$C(y) = \sin \, y .$

Hence

$f(x,y) = 2x^2y + x + \sin \, y.$

The solution to the differential equation is

$2x^2y + x + \sin \, y = C.$

Does this method always work? The answer is no. We can tell if the method works by remembering that for a function with continuous partial derivatives, the mixed partials are order independent. That is

$f_{xy} = f_{yx} .$

If we have the differential equation

$M(x,y) + N(x,y)y' = 0$

then we say it is an exact differential equation if

$M_y(x,y) = N_x(x,y) .$

Theorem (Solutions to Exact Differential Equations)

Let $$M$$, $$N$$, $$M_y$$, and $$N_x$$ be continuous with

$M_y = N_x.$

Then there is a function $$f(x,y)$$ with

$$f_x = M$$ and $$f_y = N$$

such that

$f(x,y) = C$

is a solution to the differential equation

$M(x,y) + N(x,y)y' = 0 .$

Example $$\PageIndex{2}$$

Solve the differential equation

$y + (2xy - e^{-2y})y' = 0 .$

Solution

We have

$M(x,y) = y$

and

$N(x,y) = 2xy - e^{-2y}.$

Now calculate

$M_y = 1 \;\;\; \text{and} \;\;\; N_x = 2y.$

Since they are not equal, finding a potential function $$f$$ is hopeless. However there is a glimmer of hope if we remember how we solved first order linear differential equations. We multiplied both sides by an integrating factor $$m$$. We do that here to get

$mM + mN_y' = 0 .$

For this to be exact we must have

$(mM)_y = (mN)_x .$

Using the product rule gives

$m_yM + mM_y = m_xN + mN_x .$

We now have a new differential equation that is unfortunately more difficult to solve than the original differential equation. We simplify the equation by assuming that either m is a function of only $$x$$ or only $$y$$. If it is a function of only $$x$$, then $$m_y = 0$$ and

$mM_y = m_xN + mN_x .$

Solving for $$m_x$$, we get

$m_x = \dfrac{M_y-N_x}{N}.$

If this is a function of $$y$$ only, then we will be able to find an integrating factor that involves $$y$$ only. If it is a function of only $$y$$, then $$m_x = 0$$ and

$m_yM + mM_y = mN_x .$

Solving for $$m_y$$, we get

$m_y = \dfrac{N_x-M_y}{M} m .$

If this is a function of $$y$$ only, then we will be able to find an integrating factor that involves $$y$$ only.

For our example

$m_y = \dfrac{N_x - M_y }{M} m = \dfrac{2y-1}{y} m = (2-\frac{1}{y})m .$

Separating gives

$\dfrac{dm}{m} = (2-\frac{1}{y}) \,dy.$

Integrating gives

$ln \, m = 2y - ln\, y.$

$m = e^{2y - ln\, y} = y ^{-1}e^{2y}.$

Multiplying both sides of the original differential equation by $$m$$ gives

$y(y ^{-1}e^{2y}) + (y ^{-1}e^{2y})(2xy - e^{-2y})y' = 0$

$\implies e^{2y} + (2xe^{2y} - \frac{1}{y})y' = 0 .$

Now we see that

$M_y = 2e^{2y} = N_x.$

Which tells us that the differential equation is exact. We therefore have

$f_x (x,y) = e^{2y}.$

Integrating with respect to $$x$$ gives

$f(x,y) = xe^{2y} + C(y).$

Now taking the partial derivative with respect to $$y$$ gives

$f_y(x,y) = 2xe^{2y} + C'(y) = 2xe^{2y} - \frac{1}{y} .$

So that

$C'(y) = \frac{1}{y}.$

Integrating gives

$C(y) = ln\, y.$

The final solution is

$xe^{2y} + ln\, y = 0.$