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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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mathsmindBest ResponseYou've already chosen the best response.0
\[y = t^{1}\sin4t\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[y \prime = t^{2}\sin4t + t^{1}4\cos4t\]
 one year ago

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

mathsmindBest ResponseYou've already chosen the best response.0
\[16ty =16t( t^{1}\sin4t)=16\sin4t\]
 one year ago

mathsmindBest ResponseYou've already chosen the best response.0
\[2y \prime = 2t^{2}\sin4t + t^{1}8\cos4t\]
 one year ago

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

mathsmindBest ResponseYou've already chosen the best response.0
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
 one year ago

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