Differential equations - backwards problem

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The discussion revolves around solving the initial value problem defined by the differential equation y' + (2t + 1)y = 2cos(t) with the initial condition y(0) = 2. Participants express uncertainty in finding the solution for y(t) and suggest deriving y''(t) directly from the first derivative. One user highlights that instead of solving for y(t), it is more efficient to use the expression for y' to find y''. The multiple-choice options for y''(0) include values of 2, -2, 4, 0, and -4. The focus remains on determining the correct value of y''(0) based on the given equation.
dmoney123
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OP warned about not including efforts

Homework Statement


If y=y(t) is the solution of the initial value problem

{
y'+(2t+1)y=2cos(t)

y(0)=2

then

y''(0)=?

it is a multiple choice practice problem with choices

y''(0)=2
y''(0)=-2
y''(0)=4
y''(0)=0
y''(0)=-4

Homework Equations

The Attempt at a Solution



Im really not sure how to go about solving this.

When I try to solve for y(t) i get a huge mess.
 
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Find an expression that has y''(t) in it.
[edit]... of LCKurtz link: that's interesting: someone in your class?
 
Simon Bridge said:
Find an expression that has y''(t) in it.
[edit]... of LCKurtz link: that's interesting: someone in your class?

?? Who are you addressing? Me? Not one of my classes. I retired over 10 years ago.
 
dmoney123 said:

Homework Statement


If y=y(t) is the solution of the initial value problem

{
y'+(2t+1)y=2cos(t)

y(0)=2

then

y''(0)=?

it is a multiple choice practice problem with choices

y''(0)=2
y''(0)=-2
y''(0)=4
y''(0)=0
y''(0)=-4

Homework Equations

The Attempt at a Solution



Im really not sure how to go about solving this.

When I try to solve for y(t) i get a huge mess.

Why solve at all? You have ##y'(t) = 2 \cos(t) - (2t +1) y(t)##, so you can get ##y''(t)##.
 
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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