Question

For each differential equation in Problems 1-21, find the general solution by finding the homogeneous solution and a particular solution. 1. y' = 1 2. y' + y = 1 3. y' + y = x 4. y'' = 1 5. y'' + 4y' = 1 6. y'' + 4y = 1 7. y'' + 4y' = x 8. y'' + y' - 2y = 3 - 6x 9. y'' + y = 6e^x + 3 10. y'' - y' - 2y = 6e^x 11. y'' + y' = 6 sin 2x 12. y'' + 4y' + 5y = 2e^x 13. y'' + 3y' = sin x + 2 cos x 14. y'' + 4y' + 4y = xe^-x 15. y'' - y = x sin x 16. y'' - 3y' + 2y = e^x sin x 17. y'' - 4y' + 4y = xe^2x 18. y'' + y = 12 cos^2 x 19. y'' - 4y' + 3y = 20 cos x 20. y'' - y = 8xe^x 21. y'' - 5y' + 6y = cosh x

          For each differential equation in Problems 1-21, find the general solution by finding the homogeneous solution and a particular solution.
1. y' = 1
2. y' + y = 1
3. y' + y = x
4. y'' = 1
5. y'' + 4y' = 1
6. y'' + 4y = 1
7. y'' + 4y' = x
8. y'' + y' - 2y = 3 - 6x
9. y'' + y = 6e^x + 3
10. y'' - y' - 2y = 6e^x
11. y'' + y' = 6 sin 2x
12. y'' + 4y' + 5y = 2e^x
13. y'' + 3y' = sin x + 2 cos x
14. y'' + 4y' + 4y = xe^-x
15. y'' - y = x sin x
16. y'' - 3y' + 2y = e^x sin x
17. y'' - 4y' + 4y = xe^2x
18. y'' + y = 12 cos^2 x
19. y'' - 4y' + 3y = 20 cos x
20. y'' - y = 8xe^x
21. y'' - 5y' + 6y = cosh x
        
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For each differential equation in Problems 1-21, find the general solution by finding the homogeneous solution and a particular solution.
1. y' = 1
2. y' + y = 1
3. y' + y = x
4. y” = 1
5. y” + 4y' = 1
6. y” + 4y = 1
7. y” + 4y' = x
8. y” + y' - 2y = 3 - 6x
9. y” + y = 6e^x + 3
10. y” - y' - 2y = 6e^x
11. y” + y' = 6 sin 2x
12. y” + 4y' + 5y = 2e^x
13. y” + 3y' = sin x + 2 cos x
14. y” + 4y' + 4y = xe^-x
15. y” - y = x sin x
16. y” - 3y' + 2y = e^x sin x
17. y” - 4y' + 4y = xe^2x
18. y” + y = 12 cos^2 x
19. y” - 4y' + 3y = 20 cos x
20. y” - y = 8xe^x
21. y” - 5y' + 6y = cosh x

Added by Chloe M.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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For each differential equation in Problems 1-21, find the general solution by finding the homogeneous solution and a particular solution.
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00:01 Equation we have y w h minus y j equals to x sine x auxiliary equation is m square minus 1 that equals 0 that is m square equals to 1 then m is equal to lesser minus 1 so complementary solution for homogeneous part is y c equals to c1 e power x plus c2 e .4 minus x using method of undetermined coefficients the particular solution should be of form y equals to e cos x plus p x x plus bx cos x less c sine x plus d x sinex then y dash equals to minus b sinex plus b x plus bxxx minus bxxxxx plus c c cx plus plus d sinex plus d x x then y double dash is equal to minus a x x minus a x minus b sinex minus b sinex minus px cos x minus b sine x minus b sine x minus c sine x plus d cross x minus d x x x minus d x x x x…
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