is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Exact Equations

MathJax TeX Test Page The general form of our solutions to differential equations is $$F(x,y(x)) = C$$ So, let's differentiate with respect to x. $$\dfrac{\partial F}{\partial x} + \dfrac{\partial F}{\partial y}\dfrac{dy}{dx} = 0 \to F_x\, dx + F_y\, dy = 0 $$. That's the basic idea for this post. We are given an equation in the form of $M\, dx + N\, dy = 0$, and we have to find the F such that $F_x = M$, and $F_y = N$. If it does have an F, we call that exact. How can we determine this? We must check if $$M_y = N_x$$ If not, don't worry, it still may have a solution. Imagine the differential equation $ydx + xdy = 0$. The function $F(x,y) = xy$. Now, let's multiply through by x. $xydx + x^2dy = 0$. $M_y = x,$ and $N-x = 2x$. The two aren't equal. So, we need to figure out a way to account for that.

Integrating Factor

MathJax TeX Test Page There are two ways of finding the integrating factor. One method is guess-and-check. The other method can only be done if the integrating factor is a function of only x or only y. If $\dfrac{M_y - N_x}{N}$ has only x terms, $$\rho(x) = e ^{\int \frac{M_y - N_x}{N} dx}$$ If $\dfrac{N_x - M_y}{M}$ has only y terms, $$\rho(y) = e ^{\int \frac{N_x - M_y}{M} dy}$$ Then, you know the following equation is exact $$\rho Mdx + \rho Ndy = 0$$

Find the solution

MathJax TeX Test Page If you know an equation is exact, you just have to find the function F. $$Mdx + Ndy = 0$$. So, $M = f_x$. Therefore, you must partially integrate M with respect to x. In addition, you must partially integrate N with respect to y, because $N = f_y$. After doing that, the two functions may be different, so you combine their terms. I will illustrate this with an example.


MathJax TeX Test Page $$2xydx + (y^2 - x^2)dy = 0$$ $$N_x - M_y = -2x - (2x) = -4x, \dfrac{-4x}{2xy} = \dfrac{-2}{y}$$ That has only y terms, so it works! $$\rho = e^{\int\frac{-2}{y}\,dy} = e^{-2\ln(y)} = \frac{1}{y^2}$$ Now, I multiply both parts by the integrating factor. $$\dfrac{2x}{y}dx + (1 - \dfrac{x^2}{y^2})dy = 0$$ First, I partially integrate the first part with respect to x. $$f_x = \dfrac{2x}{y}, F = \dfrac{x^2}{y} + C$$ Now, I partially integrate the second part with respect to y. $$f_y = 1 - \dfrac{x^2}{y^2}, F = \dfrac{x^2}{y} + y + C$$ So, we combine the two terms, and we get $$\dfrac{x^2}{y} + y = C \to \boxed{x^2 + y^2 = Cy}$$
David Witten

Integrating Factor

Substitution Methods