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### Exact Equations

#### Discussion

The general solution to a first order equation has 1 arbitrary constant. If we solve for that constant, we can write the general solution to a first order equation in the form $$ F(x,y) = K $$ where $F$ is some function which depends upon the equation and $K$ is the constant. From this solution, we can now try to recover the original equation. Differentiate both sides with respect to $x$ and use the chain rule for functions of several variables to obtain $$ \frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx}=0 $$ This now suggests a technique for solving first order differential equations, find a function $F$ so that the equation takes on the form above and then the solution is $F(x,y) = K$. If we start with a general equation $$ M(x,y) + N(x,y)\frac{dy}{dx} = 0 $$ we then have the question of how to find $F$. In fact, first we should check that such an $F$ exists. Comparing this equation with the previous equation, we see that we must have $$ \begin{align} \frac{\partial F}{\partial x}&=M(x,y) \\ \frac{\partial F}{\partial y}&=N(x,y) \end{align} $$ That $M$ and $N$ are the two different partials of a single function means that $[M,N]$ forms a gradient field in the terminology of Calculus III. And of course all of you remember that a necessary and sufficient condition that a vector field is locally a gradient field is that the curl of the vector field is 0. In this setup, that means there is a function $F$ satisfying the pair of equations above if and only if $$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $$ Once we have checked that an $F$ exists, it is usually a straightforward process to find $F$ by integration. The process is sufficiently lengthy though that it is worthwhile to check that an $F$ exists before trying to construct it.

#### Example

$$ 2xy+(x^2+1)\frac{dy}{dx}=0 $$ We check that $$ \frac{\partial (2xy)}{\partial y}=2x=\frac{\partial (x^2 + 1)}{\partial x} $$ so the equation is exact. So we have $$ \begin{align} \frac{\partial F}{\partial x}&=2xy \\ \frac{\partial F}{\partial y}&=x^2+1 \end{align} $$ We now integrate the first partial derivative to obtain $$ \begin{align} \int\,\partial F &= \int 2xy\,\partial x \\ F(x,y)&=x^2y+C(y) \end{align} $$ This requires a bit of explanation. We do a partial integration because we are undoing a partial differentiation. So we integrate with respect to $x$ and treat $y$ as a constant. Finally, we end up with an arbitrary function of $y$, $C(y)$, instead of the usual arbitrary constant of integration since we are treating $y$ as a constant. So from our condition that $\partial F/\partial x=2xy$ we obtain a formula for $F$. But we can obtain another formula from $\partial F/\partial y=x^2+1$. $$ \begin{align} \int\,\partial F&=\int x^2+1\,\partial y \\ F(x,y)&=x^2y+y+\tilde{C}(x) \end{align} $$ where $\tilde{C}(x)$ is an arbitrary function of $x$. We now have two different definitions of $F$ from our two different equations. $$ \begin{align} F(x,y) &= x^2y + C(y) \\ F(x,y) &= x^2y + y + \tilde{C}(x) \end{align} $$ Comparing our two formulas, we see that they agree if we set $C(y)=y$ and $\tilde{C}(x)=0$. So we end up with $F(x,y)=x^2y+y$ and our solution is $$ x^2 y + y = K $$ where $K$ is an arbitrary constant. We are now ready to give the paradigm.#### Paradigm

$(x^2+y)+(x+\cos(y))\dfrac{dy}{dx}=0$
*Step 1:* Test for exactness
$$
\frac{\partial (x^2+y)}{\partial y}=1=\frac{\partial (x+\cos(y))}{\partial x}
$$
so the equation is exact.

*Step 2:* Write the partial differential equations.
$$
\begin{align}
\frac{\partial F}{\partial x}&=x^2+y \\
\frac{\partial F}{\partial y}&=x+\cos(y)
\end{align}
$$
*Step 3:* Integrate the first partial differential equation.
$$
\begin{align}
\int \partial F&=\int x^2+y\,\partial x \\
F(x,y)&=x^3/3+xy+C(y)
\end{align}
$$
*Step 4:* Integrate the second partial differential equation.
$$
\begin{align}
\int \partial F&=\int x+\cos(y)\,\partial y \\
F(x,y)&=xy+\sin(y)+\tilde{C}(x)
\end{align}
$$
*Step 5:* Equate expressions for $F(x,y)$.
$$
\begin{align}
F(x,y)&=x^3/3+xy+C(y) \\
F(x,y)&=xy+\sin(y)+\tilde{C}(x)
\end{align} $$
So $C(y)=\sin(y)$ and $\tilde{C}(x)=x^3/3$ and
$$
F(x,y)=x^3/3+xy+\sin(y)
$$
*Step 6:* Solution is $F(x,y)=K$.
$$
x^3/3+xy+\sin(y)=K
$$

#### Example

$\displaystyle x^2\frac{dy}{dx}=x^3-2xy,\qquad y(1)=3$FIRST: Find the general solution.

Step 1: Rewriting this as $\displaystyle (2xy-x^3)+x^2\frac{dy}{dx}=0$ the test for exactness is $$ \frac{\partial (2xy-x^3)}{\partial y}=2x=\frac{\partial x^2}{\partial x} $$ so the equation is exact.

Step 2: $$\begin{align} \frac{\partial F}{\partial x}&=2xy-x^3 \\ \frac{\partial F}{\partial y}&=x^2 \end{align} $$ Step 3: $\displaystyle F(x,y)=\int 2xy-x^3\,\partial x=x^2y-x^4/4+C(y)$

Step 4: $\displaystyle F(x,y)=\int x^2\,\partial y=x^2y+\tilde{C}(x)$

Step 5: $\displaystyle F(x,y)=x^2y-x^4/4$

Step 6: $\displaystyle x^2y-x^4/4=K$

SECOND: $$\begin{align} 1^2\times3-1^4/4&=K\\11/4&=K\end{align} $$ So
$ x^2y-x^4/4=11/4$ or $y=x^2/4+(11/4)x^{-2}$. (It is easiest to wait to
find an explicit solution until after finding the arbitrary constant when
solving an *exact* initial value problem.)

You can generate additional examples of initial value problems for first order exact equations here.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett