**Unformatted text preview: **Math 415 Homework 5 3.3.23 If f , g , and h are differentiable functions then W ( fg,fh ) = f 2 W ( g,h ). By direct calculation W ( fg,fh ) = fg fh f g + fg f h + fh = fg ( f h + fh )- fh ( f g + fg ) = ff gh + f 2 gh- ff gh- f 2 g h = f 2 gh- f 2 g h = f 2 W ( g,h ) 3.5.17 Consider the initial value problem 4 y 00 + 4 y + y = 0, y (0) = 1, y (0) = 2. (a) The equation has characteristic equation 4 r 2 + 4 r + 1 = (2 r + 1) 2 , so the general solution is y ( t ) = C 1 e- t/ 2 + C 2 te- t/ 2 . Plugging in the initial values gets us C 1 = 1 and- 1 2 C 1 + C 2 = 2, or C 2 = 5 2 . And our solution is y ( t ) = e- t/ 2 + 5 2 te- t/ 2 . (b) y ( t ) = 2 e- t/ 2- 5 4 te- t/ 2 and this equals zero when t = 8 / 5 (at which point y (8 / 5) = 5 e- 4 / 5 ). To confirm that this is a maximum point we consider the second derivative y 00 =- y- 1 4 y (this is just a rewrite of the original differential equation). At t = 8 / 5, we have y 00 (8 / 5) =-- 1 4 (5 e- 4 / 5 ) < 0, so the point ( t M ,y M ) = (8...

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