When you finish this section, you should be able to:
When a first-order differential equation is separable, that is, when it can be written as \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {M(x)\,dx+N(y)\,dy=0}}} \]
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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.
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 \(R\).
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.
Gradients are discussed in Section 13.1, pp. 866-871.
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.
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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.
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.
Problem 9.
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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 \(\ x\cos y-\sin x+e^{y}=\pi +1.\)
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.
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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.