Differential equations - backwards problem

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Homework Help Overview

The discussion revolves around an initial value problem involving a first-order linear differential equation. The problem asks for the second derivative of the solution at a specific point, given the equation and initial condition.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants express uncertainty about how to solve for the function y(t) and its derivatives. Some suggest finding an expression for y''(t) directly from the given equation rather than solving for y(t) explicitly.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to derive y''(t). There is a suggestion to focus on the relationship between y', y, and the given equation to find y'' without fully solving for y.

Contextual Notes

Participants are working within the constraints of a multiple-choice format, which may influence their approach to the problem. There is also a reference to a related thread that may provide additional context or methods.

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)##.
 

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