Solve the initial value problems in Problems 31 through 40. 31. y" + 4y = 2x; y(0) = 1, y'(0) = 2 32. y" + 3y + 2y = e*; y(0) = 0, y'(0) = 3 33. y" +9y sin 2x; y(0) = 1, y'(0) = 0 34. y"+y = cos x; y(0) = 1, y'(0) = -1 35. y" - 2y + 2y = x + 1; y(0) = 3, y'(0) = 0 36. y(4) - 4y" = x²; y(0) = y'(0) = 1, y" (0) = y(3) (0) = -1 37. y(³) — 2y" + y² = 1 + xe*; y(0) = y'(0) = 0, y" (0) = 1 =
Solve the initial value problems in Problems 31 through 40. 31. y" + 4y = 2x; y(0) = 1, y'(0) = 2 32. y" + 3y + 2y = e*; y(0) = 0, y'(0) = 3 33. y" +9y sin 2x; y(0) = 1, y'(0) = 0 34. y"+y = cos x; y(0) = 1, y'(0) = -1 35. y" - 2y + 2y = x + 1; y(0) = 3, y'(0) = 0 36. y(4) - 4y" = x²; y(0) = y'(0) = 1, y" (0) = y(3) (0) = -1 37. y(³) — 2y" + y² = 1 + xe*; y(0) = y'(0) = 0, y" (0) = 1 =
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
Number 32 please.
![Solve the initial value problems in Problems 31 through 40.
31. y" + 4y
2x; y(0) = 1, y'(0) = 2
32. y" + 3y' + 2y = e*; y(0) = 0, y'(0) = 3
sin 2x; y(0) = 1, y'(0) = 0
33. y" +9y
=
=
34. y" + y = cos x; y(0) = 1, y'(0) = −1
35. y" - 2y' + 2y = x + 1; y(0) = 3, y'(0) = 0
36. y(4) — 4y" = x²; y(0) = y'(0) = 1, y"(0) = y
37. y(³) — 2y″ + y² = 1 + xe* ; y(0) = y'(0) = 0, y″(0) = 1
(³) (0) = −1
1 (3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44cf4428-d50c-4dc1-a095-73926c558c67%2F691045f9-2a01-41cd-92f7-17df3b1535a5%2Fldt5eu8_processed.png&w=3840&q=75)
Transcribed Image Text:Solve the initial value problems in Problems 31 through 40.
31. y" + 4y
2x; y(0) = 1, y'(0) = 2
32. y" + 3y' + 2y = e*; y(0) = 0, y'(0) = 3
sin 2x; y(0) = 1, y'(0) = 0
33. y" +9y
=
=
34. y" + y = cos x; y(0) = 1, y'(0) = −1
35. y" - 2y' + 2y = x + 1; y(0) = 3, y'(0) = 0
36. y(4) — 4y" = x²; y(0) = y'(0) = 1, y"(0) = y
37. y(³) — 2y″ + y² = 1 + xe* ; y(0) = y'(0) = 0, y″(0) = 1
(³) (0) = −1
1 (3
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