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.

Question

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 · Answer

$$y = t^{-1}\sin4t$$

anonymous · Answer

$$y \prime = -t^{-2}\sin4t + t^{-1}4\cos4t$$

anonymous · Answer

$$y \prime \prime = t^{-3}2\sin4t -t^{-2}8\cos4t - t^{-1}16\sin4t$$

anonymous · Answer

$$16ty =16t( t^{-1}\sin4t)=16\sin4t$$

anonymous · Answer

$$2y \prime = -2t^{-2}\sin4t + t^{-1}8\cos4t$$

anonymous · Answer

$$16ty \prime \prime = t^{-2}32\sin4t -t^{-1}128\cos4t - 256\sin4t$$

anonymous · Answer

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

anonymous · Answer

Thank you, but that does not explain how you get the other solution: y = -πt^(-1)cos(4t)