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Problem 2: Solve the initial value problem t² y(0) = 1. 2y +2' Solution: This is a separable equation, by integration y' = ²+2y=√²+0₂ t³ C. From y(0) = 1, I get C = 3. Then I express y to get √ st Note that I took the positive square root to get the correct initial value. y(t): = t³ +4 -1.

Problem 2: Solve the initial value problem t² y(0) = 1. 2y +2' Solution: This is a separable equation, by integration y' = ²+2y=√²+0₂ t³ C. From y(0) = 1, I get C = 3. Then I express y to get √ st Note that I took the positive square root to get the correct initial value. y(t): = t³ +4 -1.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem 2: Solve the initial value problem t² 2y + 2' Solution: This is a separable equation, by integration y' = y(0): = 1. y² + 2y = 1⁄²t³ + C. 3 From y(0) = 1, I get C = 3. Then I express y to get y(t) = √ √ 1/3 √√₁1²+ Note that I took the positive square root to get the correct initial value. t³ + 4 − 1.
Transcribed Image Text:Problem 2: Solve the initial value problem t² 2y + 2' Solution: This is a separable equation, by integration y' = y(0): = 1. y² + 2y = 1⁄²t³ + C. 3 From y(0) = 1, I get C = 3. Then I express y to get y(t) = √ √ 1/3 √√₁1²+ Note that I took the positive square root to get the correct initial value. t³ + 4 − 1.
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