Write 2nd order ODE as system of two 1st order ODEs

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To convert the second-order ODE d^2y/dt^2 + 5(dy/dt)^2 - 6y + e^{sin(t)} = 0 into a system of first-order ODEs, let w = dy/dt. The correct formulation is dy/dt = w and dw/dt = -5w^2 + 6y - e^{sin(t)}, which differs from the solution provided in the book. The book's solution of dw/dt = -5w - 6y + e^{sin(t)} is questioned by multiple participants. There is consensus that the book's answer may be incorrect, as others arrived at the same conclusion. Clarification on the correct formulation is sought among the participants.
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Homework Statement


Write the following second-order ODE as a system of two first-order ODEs.

##d^2y/dt^2 + 5(dy/dt)^2 - 6y + e^{sin(t)} = 0##

Homework Equations


w = dy/dt

The Attempt at a Solution


The solution of the book says ##dy/dt = w, dw/dt = -5w - 6y + e^{sin(t)}##, but shouldn't it be ##w = dy/dt, dw/dt = -5w^2 + 6y - e^{sin(t)}##, or am I missing something?
 
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I think you are missing the correct solution manual because I came to the same answer that you did ¯\_(ツ)_/¯
 
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Lol, great. :P

Thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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