Chapter 4 21. (a) With y ( t )= e rt , y0 ( t )= re rt , the equation becomes are rt + be rt =( ar + b ) e rt =0 . Since the function e rt is never zero on ( −∞ , ∞ ), to satisfy the above equation we must have ar + b =0 . (b) Solving the characteristic equation, ar + b = 0, obtained in part (a), we get r = − b/a .So y ( t )= e rt = e − bt/a , and a general solution is given by y = ce − bt/a ,where c is an arbitrary constant. 23. We form the characteristic equation, 5 r + 4 = 0, and Fnd its root r = − 4 / 5. Therefore, y ( t )= ce − 4 t/ 5 is a general solution to the given equation. 25. The characteristic equation, 6 r − 13 = 0, has the root r =13 / 6. Therefore, a general solution is given by w ( t )= ce 13 t/ 6 . 27. Assuming that y 1 ( t )= e − t cos 2 t and y 2 ( t )= e − t sin 2 t are linearly dependent on (0 , 1), we conclude that, for some constant
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.