University:
University of California, Berkeley
Course:
MATH 51 | Calculus I
Academic year:

2023
Views:

55

Pages:

1
Author:

Clint W.

Section Method of Variation of Parameters To determine a particular solution to ay ′′ + by ′ + cy = f : (a) Find two linearly independent solutions { y1 (x), y2 (x) } to the corresponding homogeneous equation and take yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x). (b) Determine u′1 (x) and u′2 (x) by using the following formulas: u′1 (x) f (x) and u′2 (x) = = − y2W y1 f (x) W , where W = y1 y2′ − y2 y1′ . Integrate to find u1 and u2 . (c) Substitute u1 (x) and u2 (x) into the expression for yp (x) to obtain a particular solution. Exercise 1. Find a general solution on (−π/2, π/2) to d2 y dt2 + y = tan t. (Nagle Section 4.6 Example 1) Exercise 2. Find a general solution on (−π/2, π/2) to d2 y dt2 + y = tan t + 3t − 1. (Nagle Section 4.6 Example 2) Exercise 3. Find a particular solution of the variable coefficient linear equation t2 y ′′ − 4ty ′ + 6y = 4t3 , t > 0, given that y1 (t) = t2 and y2 (t) = t3 are solutions to the corresponding homogeneous equation. (Nagle Section 4.6 Example 3) Exercise 4. Solve each differential equation by variation of parameters. (#4, 6) (a) y ′′ + y = sec θ tan θ (b) y ′′ + y = sec2 x Class Exercise 1. Solve each differential equation by variation of parameters. (#8, 10, 12, 14, 16, 18) ex (a) y ′′ − y = sinh 2x (b) 4y ′′ − y = ex/2 + 3 (c) y ′′ − 2y ′ + y = 1+x 2 √ (d) y ′′ − 2y ′ + y = et arctan t (e) 2y ′′ + y ′ = 6x (f) 4y ′′ − 4y ′ + y = ex/2 1 − x2 Exercise 5. Solve the differential equation 2y ′′ + y ′ − y = x + 1 by variation of parameters, subject to the initial conditions y(0) = 1, y ′ (0) = 0. (#20) Class Exercise 2. Solve the differential equation y ′′ − 4y ′ + 4y = (12x2 − 6x)e2x by variation of parameters, subject to the initial conditions y(0) = 1, y ′ (0) = 0. (#22) Exercise 6. Solve the differential equation y ′′ − 4y = e2x x by variation of parameters. (#24) √ Class Exercise 3. Solve the differential equation 2y ′′ + 2y ′ + y = 4 x by variation of parameters. (#26) For a homogeneous third-order differential equation, the particular solution is yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x) + u3 (x)y3 (x), where y1 , y2 , and y3 are solutions and u1 , u2 , and u3 are determined from u′1 = y1 W = y1′ y1′′ y2 y2′ y2′′ 0 y2 y3 y3′ , W1 = 0 y2′ ′′ y3 f (x) y2′′ W1 W , u′2 = W2 W , y3 y1 y3′ , W2 = y1′ y3′′ y1′′ u′3 = W3 W , 0 y3 y1 0 y3′ , W3 = y1′ f (x) y3′′ y1′′ y2 y2′ y2′′ 0 0 f (x) Exercise 7. Solve the third-order differential equation y ′′′ + 4y ′ = sec 2x by variation of parameters. (#30) Class Exercise 4. Solve the third-order differential equation y ′′′ − 3y ′′ + 2y ′ = of parameters. (#32) Homework: 3-31 (every 4th) e2x 1+ex by variation