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2.4 Exact Equations: Class Notes

2.4a

A partial derivative is an ordinary derivative holding all but one variable constant.

Computation and notation: for $f(x,y)=x^2{y}^3$,

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle f_x=\frac{\partial f}{\partial x}}& {\displaystyle =2x{y}^3}\hfill \\ \hfill {\displaystyle f_y=\frac{\partial f}{\partial y}}& {\displaystyle =3x^2{y}^2}\hfill \end{array}}$$

Here $f_x$ treats $y$ as a constant, and $f_y$ treats $x$ as a constant. Taking one more partial derivative:

$${\displaystyle f_{xy}=\frac{{\partial }^{2}f}{\partial y\partial x}=6x{y}^2{f}_{yx}=6x{y}^2}$$

The second mixed partial derivatives are equal: $f_{xy}=f_{yx}$.

Now, for $z=f(x,y)$,

$$dz=f_{x}dx+f_{y}dy$$

This is the full differential of $z$, or the exact differential.

Consider $f(x,y)=C$ as a level curve of a surface. For example, if $f(x,y)=x^2+y^2$, then $z=f(x,y)=C$ becomes $x^2+y^2=C$, a circle of radius $\sqrt{C}$. Each such circle is a level curve of the surface $z=x^2+y^2$:

The bowl-shaped surface $z=x^2+y^2$. Horizontal slices at $z=1$ and $z=4$ are circles; projected down to the $xy$-plane (dashed) they are the level curves $x^2+y^2=1$ and $x^2+y^2=4$.

For the level curve $f(x,y)=C$, the (full) exact differential is

$$f_{x}dx+f_{y}dy=0$$

For our work, we will consider

$$Mdx+Ndy=0,$$

the solution to which is $f(x,y)=C$. So $M$ plays the role of $f_x$ and $N$ plays the role of $f_y$:

$${\displaystyle \begin{array}{c}\hfill {\displaystyle f(x,y)}\hfill \\ \hfill {\displaystyle \swarrow \frac{\partial }{\partial x}\frac{\partial }{\partial y}\searrow }\hfill \\ \hfill {\displaystyle f_x=MN=f_y}\hfill \end{array}}$$

In a problem, $M$ and $N$ are the starting point, and we look for $f$ (for instance, ${\displaystyle f(x,y)=\int Mdx}$). Since the mixed partials of $f$ are equal, if such an $f$ exists then

$$f_{xy}=M_y=N_x=f_{yx}$$

This is the test for exactness: $Mdx+Ndy=0$ is exact when $M_y=N_x$.

2.4b

Example

Determine whether the given differential equation is exact. If it is exact, solve it.

$$(\sin y-y\sin x)dx+(\cos x+x\cos y-y)dy=0$$

Solution

If $M_y=N_x$, then $Mdx+Ndy=0$ is exact. Here

$$M=\sin y-y\sin x,N=\cos x+x\cos y-y$$

$$M_y=\cos y-\sin x,N_x=-\sin x+\cos y$$

These are equal, so the equation is exact. Integrate $M$ with respect to $x$:

$${\displaystyle f(x,y)=\int Mdx=\int (\sin y-y\sin x)dx=x\sin y+y\cos x+h(y)}$$

Tip: The "constant" of integration is a function $h(y)$, because $y$ is constant with respect to $x$.

Now use $f_y=N$: differentiate with respect to $y$ and set it equal to $N$:

$$x\cos y+\cos x+h\prime (y)=\cos x+x\cos y-y$$

The terms $x\cos y$ and $\cos x$ cancel, leaving

$${\displaystyle h\prime (y)=-y\Rightarrow h(y)=-\frac{1}{2}{y}^2+k}$$

$${\displaystyle f(x,y)=x\sin y+y\cos x-\frac{1}{2}{y}^2+k}$$

The solution has the form $f(x,y)=C$ (the constant $k$ is absorbed into $C$):

Example

$${\displaystyle (1+\ln x+\frac{y}{x})dx=(1-\ln x)dy}$$

Solution

Put the equation in the form $Mdx+Ndy=0$:

$${\displaystyle (1+\ln x+\frac{y}{x})dx+(\ln x-1)dy=0}$$

$${\displaystyle M=1+\ln x+\frac{y}{x},N=\ln x-1}$$

$${\displaystyle M_y=\frac{1}{x},N_x=\frac{1}{x}}$$

These are equal, so the equation is exact. Integrating $M$ with respect to $x$ would need integration by parts for $\ln x$, so start with $N$ instead:

$${\displaystyle f(x,y)=\int Ndy=\int (\ln x-1)dy=y\ln x-y+g(x)}$$

Now use $M=f_x$:

$${\displaystyle 1+\ln x+\frac{y}{x}=\frac{y}{x}+g\prime (x)\Rightarrow g\prime (x)=1+\ln x}$$

For ${\displaystyle \int \ln xdx}$, integrate by parts with $u=\ln x$, $dv=dx$, so ${\displaystyle du=\frac{1}{x}dx}$, $v=x$:

$${\displaystyle g(x)=\int (1+\ln x)dx=x+x\ln x-x+k=x\ln x+k}$$

The solution has the form $f(x,y)=C$:

Note: the form of the solution is not $y=\dots$ or $f(x,y)=\dots$; it is $f(x,y)=C$, an implicit solution.

2.4c

Example

Solve the given initial-value problem.

$$(e^x+y)dx+(2+x+y{e}^y)dy=0,y(0)=1$$

Solution

Here $M_y=1$ and $N_x=1$, so the equation is exact. This time, integrate both $M$ and $N$:

$${\displaystyle \text{(1)}f(x,y)=\int Mdx=\int (e^x+y)dx=e^x+xy+h(y)}$$

$${\displaystyle \text{(2)}f(x,y)=\int Ndy=\int (2+x+y{e}^y)dy=2y+xy+\int y{e}^{y}dy}$$

Integrate by parts with $u=y$, $dv=e^{y}dy$, so $du=dy$, $v=e^y$: ${\displaystyle \int y{e}^{y}dy=y{e}^y-e^y}$. So

$$f(x,y)=2y+xy+y{e}^y-e^y+g(x)$$

Comparing (1) and (2): the term $xy$ appears in both; the terms with only $y$ give $h(y)=2y+y{e}^y-e^y$, and the term with only $x$ gives $g(x)=e^x$. Combining them:

$$2y+xy+y{e}^y-e^y+e^x=C$$

Check: $f_x=y+e^x=M$ and $f_y=2+x+y{e}^y+e^y-e^y=2+x+y{e}^y=N$, so $f_{x}dx+f_{y}dy=0$ is $(y+e^x)dx+(2+x+y{e}^y)dy=0$.

Use the initial condition $y(0)=1$: when $x=0$, $y=1$:

$$2+0+e-e+1=C\Rightarrow C=3$$

Example

Solve the given differential equation by finding an appropriate integrating factor.

$${\displaystyle \cos xdx+(1+\frac{2}{y})\sin xdy=0}$$

Solution

$${\displaystyle M_y=0,N_x=(1+\frac{2}{y})\cos x}$$

These are not equal, so the equation is not exact. The integrating factor is either

$${\displaystyle \mu (x)=e^{\int \frac{M_y-N_x}{N}dx}\text{or}\mu (y)=e^{\int \frac{N_x-M_y}{M}dy}}$$

Then $\mu Mdx+\mu Ndy=0$ is exact. Try $\mu (x)$:

$${\displaystyle \frac{M_y-N_x}{N}=\frac{-(1+\frac{2}{y})\cos x}{(1+\frac{2}{y})\sin x}=-\cot x}$$

$${\displaystyle \mu (x)=e^{\int -\cot xdx}=e^{-\ln |\sin x|}=e^{\ln |\sin x{|}^{-1}}=\csc x}$$

The minus sign moves into the logarithm as the exponent $-1$. Multiply the equation by $\csc x$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \csc x\cos xdx+\csc x(1+\frac{2}{y})\sin xdy}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle \cot xdx+(1+\frac{2}{y})dy}& {\displaystyle =0}\hfill \end{array}}$$

This is exact (now $M_y=0=N_x$). Integrate $M$ with the substitution $u=\sin x$:

$${\displaystyle f(x,y)=\int \cot xdx=\int \frac{\cos x}{\sin x}dx=\int \frac{du}{u}=\ln |\sin x|+h(y)}$$

Now $f_y=N$:

$${\displaystyle h\prime (y)=1+\frac{2}{y}\Rightarrow h(y)=y+2\ln |y|+k=y+\ln y^2+k}$$

2.4d

Where does the integrating factor come from?

We want $\mu Mdx+\mu Ndy=0$ to be exact, so we need

$$\begin{array}{cc}\hfill {\displaystyle (\mu M{)}_y}& {\displaystyle =(\mu N{)}_x}\hfill \\ \hfill {\displaystyle {\mu }_{y}M+\mu M_y}& {\displaystyle ={\mu }_{x}N+\mu N_x}\hfill \\ \hfill {\displaystyle {\mu }_{y}M-{\mu }_{x}N}& {\displaystyle =(N_x-M_y)\mu }\hfill \end{array}$$

If $\mu =\mu (x,y)$ depends on both variables, this is a partial differential equation (PDE), which is not for this class.

Suppose $\mu =\mu (x)$. Then ${\mu }_y=0$, and we have

$${\displaystyle \frac{{\mu }_x}{\mu }=\frac{M_y-N_x}{N}\Rightarrow \ln \mu =\int \frac{M_y-N_x}{N}dx\Rightarrow \mu =e^{\int \frac{M_y-N_x}{N}dx}}$$

If $\mu =\mu (y)$, then ${\mu }_x=0$, and

$${\displaystyle \frac{{\mu }_y}{\mu }=\frac{N_x-M_y}{M}\Rightarrow \mu =e^{\int \frac{N_x-M_y}{M}dy}}$$