A community for students.
Here's the question you clicked on:
 0 viewing
solidknight
 2 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.
solidknight
 2 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.

This Question is Open

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

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

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

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

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

mathsmind
 2 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

solidknight
 2 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)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.