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

2023
Views:

39

Pages:

5
Author:

rezancuk2j

Solutions: Section 2.2 1. Problem 1: Give the general solution: y 0 = x2 /y y dy = x2 dx 1 2 1 3 y = x +C 2 3 ⇒ 2. Problem 3: Give the general solution to y 0 + y 2 sin(x) = 0. First write in standard form: dy = −y 2 sin(x) dx ⇒ − 1 dy = sin(x) dx y2 Before going any further, notice that we have divided by y, so we need to say that this is value as long as y(x) 6= 0. In fact, we see that the function y(x) = 0 IS a possible solution. With that restriction in mind, we proceed by integrating both sides to get: 1 = − cos(x) + C y ⇒ y= 1 C − cos(x) 3. Problem 7: Give the general solution: dy x − e−x = dx y + ey First, note that dy/dx exists as long as y 6= ey . With that requirement, we can proceed: (y + ey ) dy = x + e−x dx Integrating, we get: 1 2 1 y + ey = x2 − e−x + C 2 2 In this case, we cannot algebraically isolate y, so we’ll leave our answer in this form (we could multiply by two). 4. Problem 9: Let y 0 = (1 − 2x)y 2 , y(0) = −1/6. First, we find the solution. Before we divide by y, we should make the note that y 6= 0. We also see that y(x) = 0 is a possible solution (although NOT a solution that satisfies the initial condition). Now solve: Z y −2 dy = Z (1 − 2x) dx ⇒ −y −1 = x − x2 + C Solve for the initial value: 6=0+C ⇒C =6 The solution is (solve for y): y(x) = x2 1 1 = −x−6 (x − 3)(x + 2) The solution is valid only on −2 < x < 3, and we could plot this by hand (also see the Maple worksheet). 5. Problem 11: x dx + ye−x dy = 0, y(0) = 1 To solve, first get into a standard form, multiplying by ex , and integrate (integration by parts for the right hand side): Z y dy = − Z xex dx 1 2 y = −xex + ex + C 2 ⇒ We could solve for the constant before isolating y: 1 =0+1+C 2 C=− 1 2 Now solve for y: y 2 = 2ex (x − 1) − 1 2 and take the positive root, since y(0) = +1. y= q 2ex (1 − x) − 1 The solution exists as long as: 2ex (1 − x) − 1 ≥ 0 We use Maple to solve where this is equal to zero (see the Worksheet online). From that, we see that −1.678 ≤ x ≤ 0.768 6. Problem 14: dy = xy 3 (1 + x2 )−1/2 y(0) = 1 dx Since we’ll divide by y, we look at the case where y = 0. We see that it is a possible solution, but not for this initial value, therefore, y 6= 0: Z y −3 dy = Z √ x dx x2 + 1 To integrate the right side of the equation, let u = x2 + 1. Integrating, we get: √ 1 − y −2 = x2 + 1 + C 2 ⇒ √ 1 = C − 2 x2 + 1 2 y2 We could solve for the constant now: 1 = C2 − 2, so C = 3. Solve for y: y(x) = q 3− 1 √ x2 + 1 where we take the positive root since the initial condition was positive. The solution will exist as long as the denominator is not zero. Solving, √ √ √ x2 + 1 = 3/2 x = ± 5/2 3 − 2 x2 + 1 = 0 √ The solution is valid for − 7. Problem 16: 5 2 √ 4, we see that y(t) → 4 as t → ∞. NOTE: I would accept the above solution for y(t) on a quiz or exam, however, it is 2 better to simplify it a bit by dividing numerator and denominator by Ae(2/3)t .

Differential Equations Solutions Section 2.2

Related Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock