Solve for -3/(2t^-2) - 2t + 7/2

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The discussion centers on solving the differential equation represented by y'' = -3/(t^3). Participants clarify the integration process and correct the notation used, specifically addressing the confusion between variables 'y' and 's'. The correct form of the expression -3/(2t^-2) is identified as equivalent to (-3/2)t^(-2), which simplifies to -3/(2t^2). The importance of proper integration and variable consistency is emphasized throughout the conversation.

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y''= -3/(t^3)

s=0 when t = 1
s'= -3/2 when t= (3)^1/2

my answer is...

-3/(2t^-2) - 2t + 7/2

please tell me if that is correct
 
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Not really.You should have integrated twice correctly.

[tex]y'=-3\frac{t^{-2}}{-2}+...[/tex]

Daniel.
 
First, you have "y" in the differential equation but "s" in the initial conditions. Which is it?
Second, you have -3/(2t^-2) in your answer which would be the same as (-3/2)t^2
Do you mean (-3/2)t^(-2)= -3/(2t^2)?

What is the integral of -3/(t^3)= -3t^(-3)? What is the integral of that?
 

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