1.5 Problems Find general solutions of the differential equations in Prob- lems 1 through 25. If an initial condition is given, find the corresponding particular solution. Throughout, primes denote derivatives with respect to x. 21. xy' = 3y + x*cos x, y(27) = 0 22. y = 2xy + 3x² exp(x²), y(0) = 5 23. ху + (2х — 3)у %3D 4x4 24. (x² + 4) y' + 3xy = x, y(0) = 1 dy 25. (x² + 1) + 3x³y = 6x exp (-x²), y(0) = 1 %3D 1. у +у3D2, у(0) %3D0 2. y' – 2y = 3e2, y(0) = 0 3. у + Зу 3D 2хе-м 4. у — 2ху 3 е 5. xy +2y = 3x, y(1) = 5 6. xy' + 5y = 7x², y(2) = 5 7. 2xy' + y = 10 8. Зху + у 3D 12х 9. ху' — у 3D х, у(1) — 7 10. 2ху — Зу %3 9х3 11. ху + у %3D Зxху, у(1) %3D 0 12. xy + 3y = 2x³, y(2) = 1 13. y +y = e", y(0) = 1 14. xy' – 3y = x², y(1) = 10 15. у + 2ху %3D х, у(0) — — 2 16. у' 3 (1 — у) cos x, у(л) %3D 2 17. (1 + х)у' + у%3 cos x, у(0) %3D1 18. ху'— 2у + x* cos x 19. у + у сotx %3D cosx 20. y = 1+x + y+xy, y(0) = 0 dx Solve the differential equations in Problems 26 through 28 by regarding y as the independent variable rather than x. = 26. (1 – 4xy²)dy y3 dx dy 27. (x + ye') = 1 dx dy 28. (1+ 2xy) = 1+ y? dx 29. Express the general solution of dy/dx = 1+2xy in terms of the error function 2 erf(x) = - dt. 30. Express the solution of the initial value problem %3D у + 2х сos x, у(1) %3D 0 dx as an integral as in Example 3 of this section.
1.5 Problems Find general solutions of the differential equations in Prob- lems 1 through 25. If an initial condition is given, find the corresponding particular solution. Throughout, primes denote derivatives with respect to x. 21. xy' = 3y + x*cos x, y(27) = 0 22. y = 2xy + 3x² exp(x²), y(0) = 5 23. ху + (2х — 3)у %3D 4x4 24. (x² + 4) y' + 3xy = x, y(0) = 1 dy 25. (x² + 1) + 3x³y = 6x exp (-x²), y(0) = 1 %3D 1. у +у3D2, у(0) %3D0 2. y' – 2y = 3e2, y(0) = 0 3. у + Зу 3D 2хе-м 4. у — 2ху 3 е 5. xy +2y = 3x, y(1) = 5 6. xy' + 5y = 7x², y(2) = 5 7. 2xy' + y = 10 8. Зху + у 3D 12х 9. ху' — у 3D х, у(1) — 7 10. 2ху — Зу %3 9х3 11. ху + у %3D Зxху, у(1) %3D 0 12. xy + 3y = 2x³, y(2) = 1 13. y +y = e", y(0) = 1 14. xy' – 3y = x², y(1) = 10 15. у + 2ху %3D х, у(0) — — 2 16. у' 3 (1 — у) cos x, у(л) %3D 2 17. (1 + х)у' + у%3 cos x, у(0) %3D1 18. ху'— 2у + x* cos x 19. у + у сotx %3D cosx 20. y = 1+x + y+xy, y(0) = 0 dx Solve the differential equations in Problems 26 through 28 by regarding y as the independent variable rather than x. = 26. (1 – 4xy²)dy y3 dx dy 27. (x + ye') = 1 dx dy 28. (1+ 2xy) = 1+ y? dx 29. Express the general solution of dy/dx = 1+2xy in terms of the error function 2 erf(x) = - dt. 30. Express the solution of the initial value problem %3D у + 2х сos x, у(1) %3D 0 dx as an integral as in Example 3 of this section.
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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please send handwritten solution for Q 24
Transcribed Image Text:1.5 Problems
Find general solutions of the differential equations in Prob-
lems 1 through 25. If an initial condition is given, find the
corresponding particular solution. Throughout, primes denote
derivatives with respect to x.
21. xy' = 3y + x*cos x, y(27) = 0
22. y = 2xy + 3x² exp(x²), y(0) = 5
23. ху + (2х — 3)у %3D 4x4
24. (x² + 4) y' + 3xy = x, y(0) = 1
dy
25. (x² + 1) + 3x³y = 6x exp (-x²), y(0) = 1
%3D
1. у +у3D2, у(0) %3D0
2. y' – 2y = 3e2, y(0) = 0
3. у + Зу 3D 2хе-м
4. у — 2ху 3 е
5. xy +2y = 3x, y(1) = 5
6. xy' + 5y = 7x², y(2) = 5
7. 2xy' + y = 10
8. Зху + у 3D 12х
9. ху' — у 3D х, у(1) — 7
10. 2ху — Зу %3 9х3
11. ху + у %3D Зxху, у(1) %3D 0
12. xy + 3y = 2x³, y(2) = 1
13. y +y = e", y(0) = 1
14. xy' – 3y = x², y(1) = 10
15. у + 2ху %3D х, у(0) — — 2
16. у' 3 (1 — у) cos x, у(л) %3D 2
17. (1 + х)у' + у%3 cos x, у(0) %3D1
18. ху'— 2у + x* cos x
19. у + у сotx %3D cosx
20. y = 1+x + y+xy, y(0) = 0
dx
Solve the differential equations in Problems 26 through 28 by
regarding y as the independent variable rather than x.
=
26. (1 – 4xy²)dy
y3
dx
dy
27. (x + ye') = 1
dx
dy
28. (1+ 2xy) = 1+ y?
dx
29. Express the general solution of dy/dx = 1+2xy in terms
of the error function
2
erf(x) = -
dt.
30. Express the solution of the initial value problem
%3D у + 2х сos x, у(1) %3D 0
dx
as an integral as in Example 3 of this section.
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