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anonymous
 3 years ago
Can someone please help me, show how you find the solution:
y = πt^(1)cos(4t)
For the problem:
Given that y = t^(1)sin4t is a solution of
ty'' + 2y' +16ty =0, find and graph the solution of the equation that satisfies
y(π/8) = 0 and y'(π/8) = 32.
Thank you.
anonymous
 3 years ago
Can someone please help me, show how you find the solution: y = πt^(1)cos(4t) For the problem: Given that y = t^(1)sin4t is a solution of ty'' + 2y' +16ty =0, find and graph the solution of the equation that satisfies y(π/8) = 0 and y'(π/8) = 32. Thank you.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y \prime = t^{2}\sin4t + t^{1}4\cos4t\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y \prime \prime = t^{3}2\sin4t t^{2}8\cos4t  t^{1}16\sin4t\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[16ty =16t( t^{1}\sin4t)=16\sin4t\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[2y \prime = 2t^{2}\sin4t + t^{1}8\cos4t\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[16ty \prime \prime = t^{2}32\sin4t t^{1}128\cos4t  256\sin4t\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when you plug in pi/8 in y you get zero, and pi/8 in y' you get 32, so it does prove the facts provided

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you, but that does not explain how you get the other solution: y = πt^(1)cos(4t)
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