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Unformatted text preview: Can y = sin ( t 2 ) be a solution on an interval containing t = 0 of an equation y 00 + p ( t ) y + q ( t ) y = 0 with continuous coecients? Explain your answer. NO. We have y ( t ) = sin ( t 2 ), then y ( t ) =-2 t cos ( t 2 ). So, y (0) = 0 and y (0) = 0. Now, assume that, y ( t ) = sin ( t 2 ) is a solution of an equation y 00 + p ( t ) y + q ( t ) y = 0. Then, it should also satisfy initial conditions y (0) = 0 and y (0) = 0. But, this initial value problem has a UNIQUE solution y = 0. (check your notes). There is only one solution y = 0 and obviously sin ( t 2 ) 6 = 0 . Hence, y ( t ) = sin ( t 2 ) is NOT a solution of this dierential equation....
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