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SalasSV_08_08_ex

# SalasSV_08_08_ex - 502 CHAPTER 8 TECHNIQUES OF INTEGRATION...

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EXERCISES *8.8 Determine whether the functions satisfy the differential equation. 1. 2 y ± y = 0; y 1 ( x ) = e x / 2 , y 2 ( x ) = x 2 + 2 e x / 2 . 2. y ± + xy = x ; y 1 ( x ) = e x 2 / 2 , y 2 ( x ) = 1 + Ce x 2 / 2 , 3. y ± + y = y 2 ; y 1 ( x ) = 1 e x + 1 , y 2 ( x ) = 1 Ce x + 1 , 4. y ±± + 4 y = 0; y 1 ( x ) = 2 sin 2 x , y 2 ( x ) = 2 cos x . 5. y ±± 4 y = 0; y 1 ( x ) = e 2 x , y 2 ( x ) = C sinh 2 x , 6. y ±± 2 y ± 3 y = 7 e 3 x ; y 1 ( x ) = e x + 2 e 3 x , y 2 ( x ) = 7 4 xe 3 x . ±ind the general solution. 7. y ± 2 y = 1. 8. xy ± 2 y =− x . 9. 2 y ± + 5 y = 2. 10. y ± y 2 e x . 11. y ± 2 y = 1 2 x . 12. xy ± + 2 y = cos x x . 13. xy ± 4 y 2 nx . 14. y ± + y = 2 + 2 x . 15. y ± e x y = 0. 16. y ± y = e x . 17. (1 + e x ) y ± + y = 1. 18. xy ± + y = (1 + x ) e x . 19. y ± + 2 xy = xe x 2 . 20. xy ± y = 2 x ln x . 21. y ± + 2 x + 1 y = 0. 22. y ± + 2 x + 1 y = ( x + 1) 5 / 2 . ±ind the particular solution determined by the initial condition. 23. y ± + y = x , y (0) = 1. 24. y ± y = e 2 x , y (1) = 1. 25. y ± + y = 1 1 + e x , y (0) = e . 26. y ± + y = 1 1 + 2 e x , y (0) = e . 27. xy ± 2 y = x 3 e x , y (1) = 0. 28. xy ± + 2 y = x , y (1) 1. 29. ±ind all functions that satisfy the differential equation y ± y = y ±± y ± . HINT: Set z = y ± y . 30. ±ind the general solution of y ± + ry = 0on[0 , ) where r is a constant. (a) Show that if y is a solution and y ( a ) = 0 at some number a 0, then y ( x ) = 0 for all x . (A solution y is either identically zero or never zero.)

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8.8 DIFFERENTIAL EQUATIONS; FIRST-ORDER LINEAR EQUATIONS ± 503 (b) Show that if r < 0, then all nonzero solutions are unbounded. (c) Show that if r > 0, then all solutions have limit 0 as x →∞ . (d) Describe all solutions of the equation in the case r = 0. Exercises 31 and 32 are concerned with the Frst-order linear differential equation ( ) y ± + p ( x ) y = 0 where the function p is continuous on an interval I . 31. (a) Show that if y 1 and y 2 are solutions of ( ), then u = y 1 + y 2 is also a solution of ( ).
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SalasSV_08_08_ex - 502 CHAPTER 8 TECHNIQUES OF INTEGRATION...

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