5. (25 pts) The functions y1(t) = cos(t) and y2(t) = sin(t) are independent solu

Question

5. (25 pts) The functions y1(t) = cos(t) and y2(t) = sin(t) are independent solutions of y + y = 0. The variation of parameters formula says that a particular solution of the equation y + y = g(t), is given by yp(t) = y1(t) Z t 0 y2(s)g(s) W(y1, y2)(s) ds + y2(t) Z t 0 y1(s)g(s) W(y1, y2)(s) ds. a) Find yp(0) and y p (0). 4 First, yp(0) = y1(0) Z 0 0 y2(s)g(s) W(y1, y2)(s) ds + y2(0) Z 0 0 y1(s)g(s) W(y1, y2)(s) ds. Since the limits of integration are both 0, the integrals are zero and yp(0) = 0. Next, by the Fundamental Theorem of Calculus, d dt Z t 0 f(s) ds = f(t). Applying this result, y p (t) = y 1 (t) Z t 0 y2(s)g(s) W(y1, y2)(s) ds y1(t) y2(t)g(t) W(y1, y2)(t) +y 2 (t) Z t 0 y1(s)g(s) W(y1, y2)(s) ds + y2(t) y1(t)g(t) W(y1, y2)(t) . Evaluating at zero as above, y p (0) = 0 y1(0) y2(0)g(0) W(y1, y2)(0) + 0 + y2(0) y1(0)g(0) W(y1, y2)(0) = 0. That is yp(0) = 0 and y p (0) = 0. Find the general solution of y + y = t using b) the method of undetermined coefficients (annihilator method), A solution y will satisfy D2 (D2 + 1)y = 0, so y(t) = c1 + c2t + c3cos(t) + c4 sin(t). 5 Plugging this expression into the equation gives c1 + c2t = t, c1 = 0, c2 = 1. The general solution is y(t) = t + c3cos(t) + c4 sin(t). c) the method of variation of parameters. The Wronskian of y1(t) = cos(t) and y2(t) = sin(t) is W = y1y 2 y 1 y2 = 1. Then using the formula above, yp(t) = cos(t) Z t 0 s sin(s) ds + sin(t) Z t 0 s cos(s) ds. Integration by parts yields yp(t) = cos(t)[s cos(s) t 0 + Z t 0 cos(s) ds]+ sin(t)[s sin(s) t 0 Z t 0 sin(s) ds. = t cos2 (t) cos(t) sin(t) + tsin2 (t) + sin(t)[cos(t) 1] = t sin(t). The general solution is y(t) = t + c3cos(t) + c4 sin(t).

Explanation / Answer

Comment: The requirement expressed in the comment indicates make sense of the question. I read one and other times the redaction of the question. I understand there are two problems of differentials equations . I changed some phrases and sentences site, with the intention to separate these two questions.. I hope this kind of answer satisfies your ask.

Answer

The variation of parameters formula says that a particular solution of the equation y + y = g(t), is given by yp(t) = y1(t) Z t 0 y2(s)g(s) W(y1, y2)(s) ds + y2(t) Z t 0 y1(s)g(s) W(y1, y2)(s) ds. a) Find yp(0) and y p (0). 4 First, yp(0) = y1(0) Z 0 0 y2(s)g(s) W(y1, y2)(s) ds + y2(0) Z 0 0 y1(s)g(s) W(y1, y2)(s) ds. Since the limits of integration are both 0, the integrals are zero and yp(0) = 0. Next, by the Fundamental Theorem of Calculus, d dt Z t 0 f(s) ds = f(t). Applying this result, y p (t) = y 1 (t) Z t 0 y2(s)g(s) W(y1, y2)(s) ds y1(t) y2(t)g(t) W(y1, y2)(t) +y 2 (t) Z t 0 y1(s)g(s) W(y1, y2)(s) ds + y2(t) y1(t)g(t) W(y1, y2)(t) . Evaluating at zero as above, y p (0) = 0 y1(0) y2(0)g(0) W(y1, y2)(0) + 0 + y2(0) y1(0)g(0) W(y1, y2)(0) = 0. That is yp(0) = 0 and y p (0) = 0.

Find the general solution of y + y = t using b) the method of undetermined coefficients (annihilator method), The functions y1(t) = cos(t) and y2(t) = sin(t) are independent solutions of y + y = 0. A solution y will satisfy D2 (D2 + 1)y = 0, so y(t) = c1 + c2t + c3cos(t) + c4 sin(t). 5 Plugging this expression into the equation gives c1 + c2t = t, c1 = 0, c2 = 1. The general solution is y(t) = t + c3cos(t) + c4 sin(t). c) the method of variation of parameters. The Wronskian of y1(t) = cos(t) and y2(t) = sin(t) is W = y1y 2 y 1 y2 = 1. Then using the formula above, yp(t) = cos(t) Z t 0 s sin(s) ds + sin(t) Z t 0 s cos(s) ds. Integration by parts yields yp(t) = cos(t)[s cos(s) t 0 + Z t 0 cos(s) ds]+ sin(t)[s sin(s) t 0 Z t 0 sin(s) ds. = t cos2 (t) cos(t) sin(t) + tsin2 (t) + sin(t)[cos(t) 1] = t sin(t). The general solution is y(t) = t + c3cos(t) + c4 sin(t).

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