Lesson 7 - Exact Equations

The differential form $M (x, y) d x + N (x, y) d y$ is said to be exact in a rectangle R if there is a function $ϕ (x, y)$ (called the potential function) such that $ϕ_{x} (x, y) = M (x, y)$ and $ϕ_{y} (x, y) = N (x, y)$ for all $(x, y) \in R$ .

When we have an equation in this form, we must first test for exactness. Suppose the first partials of $M (x, y)$ and $N (x, y)$ are continuous in a rectangle $R$ . Then $M (x, y) d x + N (x, y) d y = 0$ is an exact equation if and only if $M_{y} (x, y) = N_{x} (x, y)$ for all $(x, y) \in R$ .

Example

(2 x^{3} + 3 y) d x + (3 x + y - 1) d y = 0

We test for exactness

\frac{d}{d y} (2 x^{3} + 3 y) = \frac{d}{d x} (3 x + y - 1)

3 = 3 ✓

Now that exactness has been determined, we can continue to solve by find the potential function of $M (x, y)$ and $N (x, y)$ .

For $M (x, y)$

\int (2 x^{3} + 3 y) d x

= \frac{1}{2} x^{4} + 3 y x

For $N (x, y)$

\int (3 x + y - 1) d y

= 3 x y + \frac{1}{2} y^{2} - y

Now comparing the solutions, we find that $y (x)$ is as follows:

\frac{1}{2} x^{4} + 3 y x + \frac{1}{2} x^{2} - y = C

Try for Yourself:

$(\tan y - 2) d x + (x s e x^{2} y + \frac{1}{y}) d y = 0$ , $y (0) = 1$
$(y^{2} - \frac{y}{x (x + y)} + 2) d x + (\frac{1}{x + y} + 2 y (x + 1)) d y = 0$

Next Lesson: Lesson 8 - Substitutions and Transformations for First Order Equations

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