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In Exercises 1-6, find the general solution of the differential equation specified. dy dy 1. = +2 dt dt 3. 5. dy dt 11. dy y 1+1 21 1+12y=3 9.-- +2, y(1)=3 =21². y(-2)=4 15. +1² In Exercises 7-12, solve the given initial-value problem. dy 7. =+2₁ y(0)=3 dt 1+1 13. = (sint)y +4 di dy y 4. - 4 cost dt + =-2ty+4e-² In Exercises 13-18, the differential equation is linear, and in theory, we can find its general solution using the method of integrating factors. However, since this method involves computing two integrals, in practice it is frequently impossible to reach a for- mula for the solution that is free of integrals. For these exercises, determine the general solution to the equation and express it with as few integrals as possible. 8. =1+19+41² + 16. y+4²+4r, y(1) 10 dy 10. =-2ty+4e². y(0) = 3 dt dy 3 12.-y=21³e", y(1)=0 dt t' dy 14. =r²y +4 dt dy =y+4cost²

In Exercises 1-6, find the general solution of the differential equation specified. dy dy 1. = +2 dt dt 3. 5. dy dt 11. dy y 1+1 21 1+12y=3 9.-- +2, y(1)=3 =21². y(-2)=4 15. +1² In Exercises 7-12, solve the given initial-value problem. dy 7. =+2₁ y(0)=3 dt 1+1 13. = (sint)y +4 di dy y 4. - 4 cost dt + =-2ty+4e-² In Exercises 13-18, the differential equation is linear, and in theory, we can find its general solution using the method of integrating factors. However, since this method involves computing two integrals, in practice it is frequently impossible to reach a for- mula for the solution that is free of integrals. For these exercises, determine the general solution to the equation and express it with as few integrals as possible. 8. =1+19+41² + 16. y+4²+4r, y(1) 10 dy 10. =-2ty+4e². y(0) = 3 dt dy 3 12.-y=21³e", y(1)=0 dt t' dy 14. =r²y +4 dt dy =y+4cost²

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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In Exercises 1-6, find the general solution of the differential equation specified. dy dt 3. 5. 7. 9. dt dy dt 11. dy dt dy dt 17. == dy 15. = dt y 1+t 2t 1+t2y = 3 In Exercises 7-12, solve the given initial-value problem. dy +2, y(0) = 3 dt +2, dt dy 2y = 21², dt t dy dt 1+t II +1² 13. = (sint)y +4 dy dt y y(1) = 3 y pt2 + 4 cost 2. 4. + cost 6. 8. 10. dy dt 12. dy -- dt t dy dt y(-2) = 4 In Exercises 13-18, the differential equation is linear, and in theory, we can find its general solution using the method of integrating factors. However, since this method involves computing two integrals, in practice it is frequently impossible to reach a for- mula for the solution that is free of integrals. For these exercises, determine the general solution to the equation and express it with as few integrals as possible. dy dt 2 - -2ty+4e- t5 y = t³ et 1 t +1 18. = :-2ty +4e-1², 3 y = 2t³e²t, y(1) = 0 t -1² y+4t² + 4t, y(1) = 10 dy 14. = t²y +4 dt dy y dt +3 dy 16. = y + 4 cost² dt 3 y (0) = 3 +t
Transcribed Image Text:In Exercises 1-6, find the general solution of the differential equation specified. dy dt 3. 5. 7. 9. dt dy dt 11. dy dt dy dt 17. == dy 15. = dt y 1+t 2t 1+t2y = 3 In Exercises 7-12, solve the given initial-value problem. dy +2, y(0) = 3 dt +2, dt dy 2y = 21², dt t dy dt 1+t II +1² 13. = (sint)y +4 dy dt y y(1) = 3 y pt2 + 4 cost 2. 4. + cost 6. 8. 10. dy dt 12. dy -- dt t dy dt y(-2) = 4 In Exercises 13-18, the differential equation is linear, and in theory, we can find its general solution using the method of integrating factors. However, since this method involves computing two integrals, in practice it is frequently impossible to reach a for- mula for the solution that is free of integrals. For these exercises, determine the general solution to the equation and express it with as few integrals as possible. dy dt 2 - -2ty+4e- t5 y = t³ et 1 t +1 18. = :-2ty +4e-1², 3 y = 2t³e²t, y(1) = 0 t -1² y+4t² + 4t, y(1) = 10 dy 14. = t²y +4 dt dy y dt +3 dy 16. = y + 4 cost² dt 3 y (0) = 3 +t
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