Chapter 2: First Order DE 2.6 Exact DE and Integrating Fa

 Bertina Cooper
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1 Chapter 2: First Order DE 2.6 Exact DE and Integrating Factor
2 First Order DE Recall the general form of the First Order DEs (FODE): dy dx = f(x, y) (1) (In this section x is the independent variable; not t.) This can also be written as M(x, y)+n(x, y) dy dx = 0 (2) with variety of choices of M(x, y), N(x, y).
3 Exact FODE Sometimes, we may be lucky: Suppose there is a function ψ(x, y) such that ψ (x, y) = M(x, y) and ψ x (x, y) = N(x, y). (3) y In that case, the equation (2) would reduce to dψ dx = ψ x + ψ dy y dx = M+N dy dx = 0 = ψ(x, y) = c where c is an arbitrary constant.
4 Exact FODE Therefore, any solution y = ϕ(x) of (2) would satisfy the equation ψ(x, ϕ(x)) = c. If we can find such a function ψ(x, y), DE (2) would be called an Exact DE. Question: When or how can we tell that a DE of the type (2) is exact? The Answer: Theorem 2.6.1
5 Exact FODE Sample I (Ex. 112) Theorem 2.6.1(page 96): Assume M(x, y), N(x, y), M, N y x { the rectangle R := (x, y) : Then the DE (2) is EXACT in R are continuous on α < x < β γ < y < δ } M y = N x at each (x, y) R. (4)
6 Continued That means there is a function ψ(x, y) satisfying equation (3) : ψ x = M, ψ y = N M, N satisfy equation (4). Proof. See page 97. For solving problems, it is important to follow the steps, as in the following samples or the examples in the textbook.
7 Sample I Exercise 2 Consider the DE (2x + 4y)+(2x 2y) dy dx = 0 (5) Determine, if it is exact. If yes, solve it.
8 Continued: Solution Here M = (2x + 4y) and N = 2x 2y. So, M y So, M = 4 and N x = 2 N y x. So, equation (5) is not exact. In the next Sample, I change equation (5) slightly, to make it exact.
9 Sample I: Exercise 2 (edited) Consider the DE (2+4yx)+(2x 2 2y) dy dx = 0 (6) Determine, if it is exact. If yes, solve it.
10 Exercise 2 (edited): Solution Here M = 2+4yx and N = 2x 2 2y. So, M y So, M = N y x = 4x and N x = 4x = 4x. So, equation (6) is exact.
11 Continued: Step 1 Sample I (Ex. 112) We set ψ x = M = 2+4yx, ψ y = N = 2x2 2y (7) Integrating the first one, with respect to x: ψ(x, y) = (2+4yx)dx + h(y) where h(y) is function of y, to be determined. So, ψ(x, y) = 2x + 2x 2 y + h(y) (8)
12 Continued: Step 2 Sample I (Ex. 112) Differentiating equation (8), then combining with the second part of the equation (7) ψ y = 2x2 + dh(y) dy = 2x 2 2y (9) (we h(y) to be independent of x) and we have: dh(y) dy = 2y = h(y) = y 2 (We do not need to have a constant). Combining with equation (8), we have ψ(x, y) = 2x+2x 2 y+h(y) = ψ(x, y) = 2x+2x 2 y y 2
13 Exercise 2 (edited): Answer So, the general solution to the DE (6) is ψ(x, y) = 2x + 2x 2 y y 2 = c where c is an arbitrary constant. For each value of c, we get a solution of (6).
14 y Sample I (Ex. 112) The Direction Field: y = (2 + 4 y x)/(2 x 2 2 y) x
15 Sample II Exercise 6 Consider the DE dy dx = ax by bx cy (10) Determine, if it is exact. If yes, solve it.
16 Exercise 6: Solution Rewrite DE (10): (ax by)+(bx cy) dy dx = 0 Here M = ax by and N = bx cy. So, M y So, M = N y x = b and N x = b if b = b. So, equation (10) if b = 0. With b = 0, we rewrite DE (10) as: ax cy dy dx = 0, With M = ax, N = cy. (11)
17 Exercise 6: Step 1 Sample I (Ex. 112) We set ψ x = M = ax, ψ y = N = cy (12) Integrating the first one, with respect to x: ψ(x, y) = axdx + h(y) where h(y) is function of y, to be determined. So, ψ(x, y) = ax2 2 + h(y) (13)
18 Exercise 6: Step 2 Sample I (Ex. 112) Differentiate (13), with respect to y and use (12) ψ y = dh(y) dy = cy (14) (we h(y) to be independent of x) and we have: dh(y) dy = cy = h(y) = cy2 2 (We do not need to have a constant).
19 Exercise 6: Answer Sample I (Ex. 112) From (13), we have ψ(x, y) = ax2 2 ax2 + h(y) = ψ(x, y) = 2 cy2 2 So, the general solution to the DE (10) is ψ(x, y) = ax2 2 cy2 2 = c where c is an arbitrary constant.
20 Sample III Exercise 10 Consider the DE ( y ) x + 6x +(ln x 2) dy dx = 0 x > 0 (15) Determine, if it is exact. If yes, solve it.
21 Exercise 10: Solution Here M = y + 6x and N = ln x 2. x So, M = 1 N and = 1 y x x x So, M = N So, DE (15) is exact. y x
22 Continued: Exercise 10: Step 2 We set ψ x = M = y x + 6x, ψ y = N = ln x 2 (16) Integrating the first one, with respect to x: (y ) ψ(x, y) = x + 6x dx + h(y) where h(y) is function of y, to be determined. So, ψ(x, y) = y ln x + 3x 2 + h(y) x > 0 (17)
23 Exercise 10: Step 3 Sample I (Ex. 112) Differentiate (17), with respect to y and use (12) ψ y = ln x + dh dx = ln x 2 (18) (we h(y) to be independent of x) and we have: dh(y) dy = 2 = h(y) = 2y (We do not need to have a constant).
24 Exercise 10: Answer From (17), we have ψ(x, y) = y ln x+3x 2 +h(y) = ψ(x, y) = y ln x+3x 2 2y So, the general solution to the DE (15) is ψ(x, y) = y ln x + 3x 2 2y = c where c is an arbitrary constant.
25 Sample I (Ex. 112) Consider the FODE (2) M(x, y)+n(x, y) dy dx = 0 (19) Question: As in 2.1, for some FODEs, can we find an integrating factor (IF) µ(x, y), so that when we multiply DE (20) by µ(x, y), the new equation becomes exact? µ(x, y)m(x, y)+µ(x, y)n(x, y) dy dx = 0 (20) Answer: Under some conditions, we can. We will not discuss IF in this section, any further.
26 2.6 Assignments and Homework Read Example 1 (motivational) Read Example 2, 3 (they explain the method of solution) Suggested Problems: Exercise 2.6 (page 101) Exercise 112 Homework: 2.6 See the Homework site!