16.3 Exact Differential Equations

OBJECTIVES

When you finish this section, you should be able to:

  1. Identify and solve an exact differential equation (p. 1074)

When a first-order differential equation is separable, that is, when it can be written as \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {M(x)\,dx+N(y)\,dy=0}}} \]

1073

a solution can be found by integrating. Under certain conditions, this idea can be extended to differential equations of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {M(x,y)\,dx+N(x,y)\,dy=0}}} \]

in which separation of variables is not possible.

spanDEFINITIONspan Exact Differential Equation

The differential equation \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {M(x,y)\,dx+N(x,y)\,dy=0}}} \]

is called an exact differential equation in a simply connected, open set \(R\) if there is a function \(z=f( x,y)\) of two variables with continuous partial derivatives on \(R\) for which \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {dz=M(x,y)\,dx+N(x,y)\,dy}}} \]

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

NEED TO REVIEW?

The differential of \(z=f(x,y)\) is discussed in Section 12.4, p. 843.

If the differential equation \(M(x,y)\,dx\,{+}\,N(x,y)\,dy=0\) is exact, then the differential of \(z=f( x,y)\) is given by \[\begin{equation*} dz=\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y} dy=M(x,y)\,dx+N(x,y)\,dy \end{equation*}\]

In other words, \(M(x,y)\,dx+N(x,y)\,dy=0\) is an exact differential equation if there is a potential function \(z=f( x,y)\) whose gradient is \(\ {\mathbf\nabla}\!f=M(x,y)\mathbf{i}+N(x,y)\mathbf{j}\). The condition for this is repeated below.

NEED TO REVIEW?

Gradients are discussed in Section 13.1, pp. 866-871.

THEOREM Test for Exactness

A differential equation of the form \[ M(x,y)\,dx+N(x,y)\,dy=0 \]

is an exact differential equation at every point \((x,y)\) in a simply connected open set \(R\) if and only if \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}}}} \]

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

So, if \(\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x},\) then the differential equation \(M(x,y)\,dx+N(x,y)\,dy=0\) is exact, and it can be written in the equivalent form \(\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy=0\), where \(z=f( x,y)\) is a function of two variables with the property that \(\dfrac{\partial f}{\partial x}=M\) and \(\dfrac{\partial f}{\partial y}=N\). But \(\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy\) is the total differential of \( z=f( x,y)\), and so \(f_{x}\,dx+f_{y}\,dy=0\) is equivalent to \( dz=0\), whose solution is \(f(x,y)=C,\) \(C\) a constant.

1 Identify and Solve an Exact Differential Equation

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Identifying and Solving an Exact Differential Equation

  1. Show that the differential equation \[\begin{equation*} (2x+3x^{2}y)\,dx+(x^{3}+2y-3y^{2})\,dy=0 \tag{1} \end{equation*}\] is an exact differential equation.
  2. Find the general solution.

Solution(a) Let \(\ M=2x+3x^{2}y\) and \(N=x^{3}+2y-3y^{2}.\) Then \[ \frac{\partial M}{\partial y}=3x^{2}\qquad \hbox{and}\qquad \frac{\partial N }{\partial x}=3x^{2} \]

Since \(\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}\) for all \(x\) and \(y\), the differential equation (1) is exact over the \(xy\)-plane.

NEED TO REVIEW?

Finding a potential function \(f\) for a conservative vector field \(F\) is discussed in Section 15.3, pp. 995-996.

(b) Since the differential equation \((2x+3x^{2}y)\,dx +(x^{3}+2y-3y^{2})\,dy=0\) is exact, there is a function \(z=f( x,y)\) so that \[ \frac{\partial f}{\partial x}=M=2x+3x^{2}y\qquad \hbox{and}\qquad \frac{\partial f}{\partial y}=N=x^{3}+2y-3y^{2} \]

We find \(f\) by integrating \(\dfrac{\partial f}{\partial x}=M=2x+3x^{2}y\) partially with respect to \(x\) (holding \(y\) constant). Then \[ \begin{equation*} f( x,y) =\int ( 2x+3x^{2}y)\,dx=x^{2}+x^{3}y+B(y) \tag{2} \end{equation*}\]

where the constant of integration is a function \(B\) of \(y\) alone, which is as yet unknown. To determine \(B(y)\), we use the fact that \(f\) must also satisfy \(\dfrac{\partial f}{\partial y}=N=x^{3}+2y-3y^{2}\). Then from (2), \[\begin{equation*} \frac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}[ x^{2}+x^{3}y+B(y)] =x^{3}+\dfrac{\partial }{\partial y}B(y)\qquad \hbox{and}\qquad \frac{\partial f}{\partial y}=x^{3}+2y-3y^{2} \end{equation*}\]

Equating the two expressions, we have \[\begin{eqnarray*} x^{3}+\dfrac{\partial }{\partial y}B(y) &=&x^{3}+2y-3y^{2} \\[4pt] \dfrac{\it dB}{dy} &=&2y-3y^{2}\qquad \color{#0066A7}{{B\hbox{ is a function of }y\hbox{ alone; }\quad \dfrac{\partial B}{\partial y}=\dfrac{dB}{dy}.}} \end{eqnarray*}\]

Now we integrate \(\dfrac{\it dB}{dy}=2y-3y^{2}\) with respect to \(y.\) \[\begin{equation*} B(y) =\int ( 2y-3y^{2})\,dy=y^{2}-y^{3}+C \end{equation*}\]

Substituting for \(B(y)\) in (2), we obtain \[\begin{equation*} f(x,y)=x^{2}+x^{3}y+y^{2}-y^{3}+C \end{equation*}\]

The general solution of the differential equation is \[\begin{equation*} x^{2}+x^{3}y+y^{2}-y^{3}+C=0 \end{equation*}\]

where \(C\) is a constant.

NOW WORK

Problem 9.

1075

Identifying and Solving an Exact Differential Equation

  1. Show that \[\begin{equation*} (\cos y-\cos x)\,dx+(e^{y}-x\sin y)\,dy=0 \end{equation*}\] is an exact differential equation.
  2. Find the particular solution that satisfies the boundary condition \(y(\pi )=0\).

Solution(a) Let \(M=\cos y-\cos x\) and \(N=e^{y}-x\sin y\). Then \[\begin{equation*} \frac{\partial M}{\partial y}=-\sin y\qquad \hbox{and}\qquad \frac{\partial N}{\partial x}=-\sin y \end{equation*}\]

Since \(\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}\) for all \(x\) and \(y\), the differential equation is exact over the \(xy\)-plane.

(b) Since the differential equation is exact, there is a function \( z=\) \(f(x,y)\) so that \[\begin{equation*} \frac{\partial f}{\partial x}=M=\cos y-\cos x\qquad \hbox{and}\qquad \frac{\partial f}{\partial y}=N=e^{y}-x\sin y \end{equation*}\]

We integrate \(\dfrac{\partial f}{\partial x}=M=\cos y-\cos x\) partially with respect to \(x\) (holding \(y\) constant). Then \[\begin{equation*} f(x,y)=\int ( \cos y-\cos x)\,dx=x\cos y-\sin x+B(y) \tag{3} \end{equation*}\]

where \(B,\) which is yet unknown, is a function of \(y\) alone. To find \(B,\) we differentiate \(f\) with respect to \(y\). \[\begin{equation*} \frac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}\left[ x\cos y-\sin x+B(y)\right] =-x\sin y-0+\dfrac{\partial }{\partial y}B(y)=-x\sin y+ \dfrac{\it dB}{dy} \end{equation*}\]

Since \(\dfrac{\partial f}{\partial y}=N=e^{y}-x\sin y,\) we have \[\begin{eqnarray*} -x\sin y+\dfrac{\it dB}{dy} &=&e^{y}-x\sin y \\ \dfrac{\it dB}{dy} &=&e^{y}\\ B(y)&=&e^{y}+C \end{eqnarray*}\]

where \(C\) is a constant. Now we substitute \(B(y)\) into (3) to obtain the general solution \[ f(x,y)=x\cos y-\sin x+e^{y}+C=0 \]

To find the particular solution that satisfies the boundary condition \(y(\pi )=0\), we need to find \(C\) so that \(y=0\) when \(x=\pi .\) Then \[\begin{eqnarray*} \pi \cos 0-\sin \pi +e^{0}+C& =&0 \\[3pt] \pi +1+C& =&0 \\[3pt] C& =&-( 1+\pi) \end{eqnarray*}\]

The particular solution is \(\ x\cos y-\sin x+e^{y}=\pi +1.\)

NOW WORK

Problem 25.

Integrating Factors

There are differential equations that are not exact, but can be made exact by multiplying the equation by an appropriate expression called an integrating factor.

1076

Transforming a Differential Equation into an Exact Differential Equation

The equation \[\begin{equation*} -y\,dx+x\,dy=0 \end{equation*}\]

is not an exact differential equation because \[\begin{equation*} \frac{\partial }{\partial y}(-y)=-1\qquad \hbox{and}\qquad \frac{\partial }{ \partial x}(x)=1 \end{equation*}\]

are not equal. The differential equation, however, can be converted into an exact differential equation by multiplying by \(\dfrac{1}{x^{2}}\). Then \[\begin{equation*} -\frac{y}{x^{2}}\,dx+\frac{1}{x}\,dy=0 \end{equation*}\]

is an exact differential equation because \[ \frac{\partial }{\partial y}\left( -\frac{y}{x^{2}}\right) =-\dfrac{1}{x^{2}}\qquad \hbox{and}\qquad \frac{\partial }{\partial x}\left( \frac{1}{x}\right) =-\frac{1}{x^{2}} \]

Multiplying the differential equation in Example 3 by \(\dfrac{1}{x^{2}}\) converted the original differential equation into an exact differential equation. An expression that makes a differential equation exact is called an integrating factor. Problems 27 and 28 discuss two situations for which an integrating factor can be found for a first-order differential equation.