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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 C e 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 . Find 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 = x e 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 . Find 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 e x , y (1) = − 1. 29. Find all functions that satisfy the differential equation y y = y y . HINT: Set z = y y . 30. Find the general solution of y + ry = 0 on [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 first-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 ( ). (b) Show that if y is a solution of ( ) and C is a constant, then u = Cy is also a solution of ( ).
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