In Exercises 12-15, use the given general solution to find a so- lution of the differential equation having the given initial con- dition. Sketch the solution, the initial condition, and discuss the solution's interval of existence. 12. y'+4y = cost, y(t) = (4/17) cost+(1/17) sint+Ce-4, y(0) = -1 13. ty' + y², y(t) = (1/3)² + C/t, y(1) = 2 14. ty' (t+1)y = 2te, y(t) = e(t + C/t), y(1) = 1/e 15. y'y(2+ y), y(t) = 2/(-1+ Ce-2), y(0) = -3
In Exercises 12-15, use the given general solution to find a so- lution of the differential equation having the given initial con- dition. Sketch the solution, the initial condition, and discuss the solution's interval of existence. 12. y'+4y = cost, y(t) = (4/17) cost+(1/17) sint+Ce-4, y(0) = -1 13. ty' + y², y(t) = (1/3)² + C/t, y(1) = 2 14. ty' (t+1)y = 2te, y(t) = e(t + C/t), y(1) = 1/e 15. y'y(2+ y), y(t) = 2/(-1+ Ce-2), y(0) = -3
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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![In Exercises 12-15, use the given general solution to find a so-
lution of the differential equation having the given initial con-
dition. Sketch the solution, the initial condition, and discuss
the solution's interval of existence.
12. y'+4y = cost, y(t) = (4/17) cost+(1/17) sint+Ce-4,
y(0) = -1
13. ty' + y², y(t) = (1/3)² + C/t, y(1) = 2
14. ty' (t+1)y = 2te, y(t) = e(t + C/t), y(1) = 1/e
15. y'y(2+ y), y(t) = 2/(-1+ Ce-2), y(0) = -3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1c820b6-43d4-4c62-839d-ef57ab178349%2F9257fe86-7f0d-40ad-a712-d2f9ab22c880%2Fzxqpih_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In Exercises 12-15, use the given general solution to find a so-
lution of the differential equation having the given initial con-
dition. Sketch the solution, the initial condition, and discuss
the solution's interval of existence.
12. y'+4y = cost, y(t) = (4/17) cost+(1/17) sint+Ce-4,
y(0) = -1
13. ty' + y², y(t) = (1/3)² + C/t, y(1) = 2
14. ty' (t+1)y = 2te, y(t) = e(t + C/t), y(1) = 1/e
15. y'y(2+ y), y(t) = 2/(-1+ Ce-2), y(0) = -3
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