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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.

Differential Equations
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\[y = t^{-1}\sin4t\]
\[y \prime = -t^{-2}\sin4t + t^{-1}4\cos4t\]
\[y \prime \prime = t^{-3}2\sin4t -t^{-2}8\cos4t - t^{-1}16\sin4t\]

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Other answers:

\[16ty =16t( t^{-1}\sin4t)=16\sin4t\]
\[2y \prime = -2t^{-2}\sin4t + t^{-1}8\cos4t\]
\[16ty \prime \prime = t^{-2}32\sin4t -t^{-1}128\cos4t - 256\sin4t\]
when 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
Thank you, but that does not explain how you get the other solution: y = -πt^(-1)cos(4t)

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