3.25 - Math 334 Lecture #11 3.2: Linear Second Order ODEs,...

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Math 334 Lecture #11 § 3.2: Linear Second Order ODE’s, Part II Examples. (a) Recall that the functions y 1 = e - t and y 2 = e - 3 t for t R are solutions of y 00 + 4 y 0 + 3 y = 0. The Wronskian of these two solutions is W ( y 1 , y 2 )( t ) = ± ± ± ± e - t e - 3 t - e - t - 3 e - 3 t ± ± ± ± = - 3 e - 4 t + e - 4 t = - 2 e - 4 t 6 = 0 , so that y = c 1 e - t + c 2 e - 3 t is the general solution. (b) The function y 1 = x , y 2 = xe x are solutions of x 2 y 00 - x ( x + 2) y 0 + ( x + 2) y = 0 on the open interval x > 0, for which W ( y 1 , y 2 )( x ) = ± ± ± ± x xe x 1 e x + xe x ± ± ± ± = xe x + x 2 e x - xe x = x 2 e x 6 = 0 for x > 0 , so that y = c 1 x + c 2 xe x is the general solution. (c) The functions y 1 ( t ) = e 3 t and y 2 ( t ) = e 3( t - 2) are solutions of y 00 - 6 y 0 + 9 y = 0. Their Wronskian is W ( y 1 , y 2 )( t ) = ± ± ± ± e 3 t e 3( t - 2) 3 e 3 t 3 e 3( t - 2) ± ± ± ± = 3 e 6 t - 6 - 3 e 6 t - 6 = 0 for all t I. So y 1 and y 2 do not form a fundamental set of solutions. Could it be that the Wronskian of two solutions of a second order linear homogeneous
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3.25 - Math 334 Lecture #11 3.2: Linear Second Order ODEs,...

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