Why Must t Be Greater Than Zero in This Differential Equation?

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SUMMARY

The differential equation ty' + 2y = 4t^2 with the initial condition y(1) = 2 requires that t be greater than zero for valid solutions. The solution is y = t^2 + t^(-2), which is defined for t > 0. The necessity for t > 0 arises from the physical interpretation of t as time, where negative values do not apply. Additionally, while the mathematical form allows for negative t, the context of the problem restricts t to positive values to maintain relevance in real-world applications.

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Homework Statement



1. ty'+2y = 4t^2 , y(1) = 2

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The Attempt at a Solution



1. I know how to get the answer but i have a trivial question. The answer is y=t^2 + t^(-2) and t>0. I do not get why t>0. Why can't t be R except 0?
 
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I think it can be all the reals except zero, but the fact that both t's are to the power of two cancels any negatives.
So you would really just end up with all the positive reals as any negatives would produce the same solution as positives.
 
Agree with Daniiel, you can also consider it from the physical standpoint, that generally t represents time and for modelling purposes we typically restrict t \geq 0 to correspond with this notion.
 

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