Mathwizurd.com 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 $$M_y = N_x$$ This is because for some function F, $F_{xy} = F_{yx}$ by Clairaut's Theorem. $M = F_x$, so the above equation holds. However, even if this fails, the differential equation may still 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 The way to find this special integrating factor can be to assume that this differential equation has a solution: $$\rho M(x,y)dx + \rho N(x,y)dy = 0$$ This is only true when $\left(\rho M\right)_y = \left(\rho N\right)_x$, by the same logic as before. We use the multivariable chain rule to write this out: $$M\rho_y + \rho M_y = N\rho_x + \rho N_x$$ We cannot solve this. This is a difficult partial differential equation. However, we can split this up into cases. $$\text{Case 1: } \rho \text{ is only in terms of x}$$ $$M\rho_y + \rho M_y = N\rho_x + \rho N_x \to \rho M_y = N\dfrac{d\rho}{dx} + \rho N_x$$ $$\int \dfrac{M_y - N_x}{N}\,\mathrm{d}x = \int\dfrac{1}{\rho}\, \mathrm{d}\rho \to \rho = e^{\int \frac{M_y - N_x}{N}\,\mathrm{d}x}$$ Note that we stipulated that $\rho$ could only have x terms. Therefore, the right side of the equation must also only have x terms. This is only true when $\dfrac{M_y - N_x}{N}$ has only x terms. $$\text{Case 2: } \rho \text{ has only y terms}$$ We do the same procedure, eliminating the $\rho_x$ term, and we end up getting $$\rho = e^{\int \frac{N_x - M_y}{M}\, \mathrm{d}y}$$ We know both of these $\rho$ scenarios satisfy $\rho M(x,y)dx + \rho N(x,y)dy = 0$. They were just special cases.

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.

Example 1

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 $$\dfrac{x^2}{y} + y = C \to \boxed{x^2 + y^2 = Cy}$$

Example 2

$$(3x^2 + y)dx + (x^2y - x)dy = 0$$ We immediately calculate $M_y$ and $N_x$. $$M_y = 1, N_x = 2xy - 1$$ Something useful is to just subtract one from the other. It doesn't matter whether it's $M_y - N_x$ or vice-versa, because you can just negate if you see that you need to divide by M.

$$M_y - N_x = 2 - 2xy$$ We can look at both equations to see if this is a multiple of either M or N. We see that N equals (-x/2) * the expression above. $$\dfrac{2 - 2xy}{x^2y - x} = \dfrac{-2(xy - 1)}{x(xy - 1)} = \dfrac{-2}{x}$$ This is written only in terms of x, so we can write $$\rho = e^{\int \frac{-2}{x}\, \mathrm{d}x} =\dfrac{1}{x^2}$$ $$\rho Mdx + \rho Ndy = 0$$ $$(3 + \dfrac{y}{x^2})dx + (y - \frac{1}{x})dy = 0$$ We know $M_y$ = $N_x$ now, so we can say $$F_x = 3 + \dfrac{y}{x^2} \to F = 3x - \frac{y}{x} + G(y)$$ $$F_y = y - \dfrac{1}{x} \to F = \dfrac{y^2}{2} - \frac{y}{x} + H(x)$$ $$F = 3x - \dfrac{y}{x} + \dfrac{y^2}{2}$$ $$\boxed{3x - \dfrac{y}{x} + \dfrac{y^2}{2} = C}$$
David Witten

Integrating Factor

Substitution Methods