Elementary Differential Equations

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 $$ M(x,y) + N(x,y)\frac{dy}{dx} = 0 \text{, or equivalently, } M(x,y)dx + N(x,y)dy = 0 $$ we then have the question of how to find $F$. In fact, first we should check that such an $F$ exists, since it would be annoying to spend time computing two different integrals and then failing at the needed algebra because no such $F$ 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 $$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $$ (that this is the same as saying the two mixed partials $\frac{\partial^2 F}{\partial x \partial y}$ and $\frac{\partial^2 F}{\partial y \partial x}$ are equal). 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 1

$$ 2xy+(x^2+1)\frac{dy}{dx}=0 $$ 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 two partial differential equations. $$ \begin{align} \int \partial F&=\int x^2+y\,\partial x = x^3/3+xy+C(y) \\ \int \partial F&=\int x+\cos(y)\,\partial y =xy+\sin(y)+\tilde{C}(x) \end{align} $$ Step 4: Equate the two expressions for $F(x,y)$. $$ x^3/3+xy+C(y) = xy+\sin(y)+\tilde{C}(x) $$ So $C(y)=\sin(y)$ and $\tilde{C}(x)=x^3/3$ and $$ F(x,y)=x^3/3+xy+\sin(y) $$ Step 5: Solution is $F(x,y)=K$. $$ x^3/3+xy+\sin(y)=K $$

Example 2: Initial Value Problem

$\displaystyle x^2\frac{dy}{dx}=x^3-2xy,\qquad y(1)=3$

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

Step 4: We find $C(y)=0$ and $\tilde{C}(x)=-x^4/4$, so we have $$F(x,y)=x^2y-x^4/4.$$

Step 5: $\displaystyle x^2y-x^4/4=K$

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.)

Example 3: Separable Equations

Any separable equation becomes an exact equation when the variables are separated. Consider $$ \frac{dy}{dx}=e^{x-y}. $$ We separate the variables, but instead of writing them on opposite sides, we put both the $x$ and the $y$ terms on the same side. $$\begin{align} \frac{dy}{dx}&=e^{x-y} \\ e^y dy &= e^x dx \\ e^x dx - e^y dy = 0. \end{align}$$ Since $\frac{\partial e^x}{\partial y} = 0 = \frac{\partial (-e^y)}{\partial x}$, this is an exact equation. We can now find the solution using the exact paradigm (starting with Step 2 since we just carried out Step 1 above).
Step 2: $$\begin{align} \frac{\partial F}{\partial x} &= e^x \\ \frac{\partial F}{\partial y} &= -e^y \\ \end{align}$$ Step 3: $$\begin{align} F(x,y)&=\int e^x \, \partial x = e^x + C(y) \\ F(x,y)&=\int -e^y \, \partial y = -e^y + \tilde{C}(x) \end{align}$$ Step 4: Equating the two expressions for $F(x,y)$, we find $C(y)=-e^y$ and $\tilde{C}(x)=e^x$ and so $$ F(x,y) = e^x - e^y. $$ Step 5: Our solution is therefore $e^x - e^y = K$.

So we have found a longer and more complicated way to solve separable equations. You might reasonably be wondering why we bothered to do this. The reason is that this is the justification for the separable paradigm we introduced earlier. You might have been concerned about why we were allowed to take apart the notation $dy/dx$ and treat it as two differentials instead of one derivative in the separable paradigm. That should have seemed (and was) unjustified at the time. But if you compare the separable paradigm with the process above, you should recognize that the separable paradigm is just taking a few shortcuts in the exact paradigm for the special case of a separable equation, which explains why we are allowed to play fast and loose with the notation.

Practice Problems

A randomly generated practice problem is below. Once you submit an answer, you will have a link to see the detailed solution for the problem, as well as the option to generate a new problem. Note that you can just quickly enter a "dummy" answer like $y=x$ to quickly get to the detailed solution if that is all you want to see.

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