University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2022
Views:

336

Pages:

5
Author:

Everett Mcknight

Solutions: Section 2.6 The material beginning on page 98, Integrating Factors for Exact Equations, will not be on the exam. 1. Problem 1: (2x + 3) + (2y − 2)y 0 = 0 We want fx = M (x, y) = 2x+3 and fy = N (x, y) = 2y −2. We check if this is possible: My = 0 Nx = 0 Now antidifferentiate M with respect to x: f (x, y) = Z M (x, y) dx = Z 2x + 3 dx = x2 + 3x + g(y) where g is some unknown function of y. Two ways of proceeding (which are equivalent). I’ll list both methods for this problem: • Try to get f from N , and compare: f (x, y) = Z N (x, y) dy = Z 2y − 2 dy = y 2 − 2y + ĝ(x) where ĝ is an unknown function of x. Comparing this to what we had before, we see that: f (x, y) = x2 + 3x + y 2 − 2y so the implicit solution is: x2 + 3x + y 2 − 2y = C NOTE: You can always check your answer! • Another method: Starting from where we left off, f (x, y) = x2 + 3x + g(y) we can see what g needs to be in order for fy = N , or: fy = g 0 (y) = 2y − 2 = N In that case, g(y) = y 2 − 2y, and f (x, y) = x2 + 3x + y 2 − 2y. The implicit solution is: x2 + 3x + y 2 − 2y = C 2. Problem 3: (3x2 − 2xy + 2) dx + (6y 2 − x2 + 3) dy = 0 Check to see if the equation is exact: My = −2x Nx = −2x So yes. Now we’ll antidifferentiate M with respect to x: f (x, y) = Z M dx = Z 3x2 − 2xy + 2 dx = x3 − x2 y + 2x + g(y) Check to see if fy is equal to N : fy = −x2 + g 0 (y) = 6y 2 − x2 + 3 so that g 0 (y) = 6y 2 + 3. That gives g(y) = 2y 3 + 3y. Put this back in to get the full solution, f (x, y) = c: x3 − x2 y + 2x + 2y 3 + 3y = C dy 3. Problem 4: (2xy 2 + 2y) + (2x2 y + 2x) dx =0 Check for “exactness”: My = 4xy + 2 Now set: f (x, y) = Z M dx = Z Nx = 4xy + 2 2xy 2 + 2y dx = x2 y 2 + 2xy + g(y) And check to see that fy = N : fy = 2x2 y + 2x + g 0 (y) = 2x2 y + 2x In this case, g 0 (y) = 0, and we don’t need to add g(y). The implicit solution: x2 y 2 + 2xy = C 4. Problem 13: (2x − y) dx + (2y − x) dy = 0 Check first: My = −1 = Nx , so the DE is exact. Now, f (x, y) = Z M dx = Z 2x − y dx = x2 − xy + g(y) where we check fy to make it equal to N (x, y): fy = −x + g 0 (y) = 2y − x g 0 (y) = 2y ⇒ ⇒ g(y) = y 2 Our implicit solution is: x2 − xy + y 2 = C With the initial condition y(1) = 3, we get: 12 − (1)(3) + 32 = C ⇒ C=7 The solution to the IVP is: x2 − xy + y 2 = 7 Now we can isolate y and find the restrictions on x (think quadratic formula in y, parentheses added for emphasis): y 2 + (−x) y + x2 − 7 = 0 Now, y= x± q x2 − 4(x2 − 7) = 2 Take the positive root since y(1) = 3. x± √ 28 − 3x2 2 The restriction on x would be that 28 − 3x2 ≥ 0. Therefore, s 28

Differential Equations Solutions: Section 2.6 - Answer Key

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