Differentials; Exact Equations MATH 374 Lecture 6 Differentials; Exact Equations

2.4: Exact Equations Definition: The differential equation M(x,y)dx + N(x,y)dy = 0 (1) is said to be exact on rectangle R if there exists a function F(x,y) such that  F/ x = M and  F/ y = N (2) on R (i.e. the total differential of F = LHS of (1) on R). 2

Note If (1) is exact, then solutions to (1) are given implicitly by F(x,y) = C where F is a function satisfying (2) and C is any constant. To see this, take the total differential of both sides of F(x,y) = C. F/x dx + F/y dy = 0 (C is a constant) M dx + N dy = 0. (by (2)) 3

A Way to Check if (1) is Exact! Theorem 2.3: If M, N, M/y, and N/x are continuous functions on rectangle R, then (1) is exact if and only if on R. 4

Proof of Theorem 2.3 (=>) Assume (1) is exact, to show (3) holds. By definition of exact, there exists a function F(x,y) such that (2) holds on R, i.e.

Proof of Theorem 2.3 (continued) (=> continued) Since M/y and N/x are continuous on R, and (3) follows. (<=) Not in text – see link on class web page for proof. 

Example 1 We now look at an example! Keep in mind that F/x = M and F/y = N must hold! (Definition of exact!) Solve (1+y2+xy2)dx + (x2y+y+2xy)dy = 0 (8) Solution: Note that the techniques we’ve learned before this section will not work. Check if (8) is exact via Theorem 2.3.

}= Example 1 (continued) Therefore (8) is exact. Hence solutions to (8) will be of the form F(x,y) = C, where

Example 1 (continued) Integrate (9) with respect to x:

Example 1 (continued) Differentiate (11) with respect to y and use (10): (10)