nagle_differential_equations_ISM_Part23

# nagle_differential_equations_ISM_Part23 - Chapter 4 20 The...

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Chapter 4 20. The auxiliary equation for this differential equation, r 2 - 4 r + 4 = ( r - 2) 2 = 0, has a double root r = 2. Thus, two linearly independent solutions are y 1 ( t ) = e 2 t and y 2 ( t ) = te 2 t . This means that a general solution is given by y ( t ) = ( c 1 + c 2 t ) e 2 t . Substituting the initial conditions into the general solution and its derivative yields y (1) = ( c 1 + c 2 t ) e 2 t | t =1 = ( c 1 + c 2 ) e 2 = 1 y (1) = ( c 2 + 2 c 1 + 2 c 2 t ) e 2 t | t =1 = (2 c 1 + 3 c 2 ) e 2 = 1 . So, c 1 = 2 e - 2 and c 2 = - e - 2 . Therefore, the solution is y ( t ) = ( 2 e - 2 - e - 2 t ) e 2 t = (2 - t ) e 2 t - 2 . 22. We substitute y = e rt into the given equation and get 3 re rt - 7 e rt = (3 r - 7) e rt = 0 . Therefore, 3 r - 7 = 0 r = 7 3 , and a general solution to the given differential equation is y ( t ) = ce 7 t/ 3 , where c is an arbitrary constant. 24. Similarly to the previous problem, we find the characteristic equation, 3 r +11 = 0, which has the root r = - 11 / 3. Therefore, a general solution is given by z ( t ) = ce - 11 t/ 3 . 26. (a) Substituting boundary conditions into y ( t ) = c 1 cos t + c 2 sin t yields 2 = y (0) = c 1 0 = y ( π/ 2) = c 2 . Thus, c 1 and c 2 are determined uniquely, and so the given boundary value problem has a unique solution y = 2 cos t . (b) Similarly to part (a), we obtain a system to determine c 1 and c 2 . 2 = y (0) = c 1 0 = y ( π ) = - c 1 . However, this system is inconsistent, and so there is no solution satisfying given boundary conditions. 106

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Exercises 4.2 (c) This time, we come up with a system 2 = y (0) = c 1 - 2 = y ( π ) = - c 1 , which has infinitely many solutions given by c 1 = 2 and c 2 – arbitrary. Thus, the boundary value problem has infinitely many solutions of the form y = 2 cos t + c 2 sin t .
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