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

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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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