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

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The discussion revolves around solving the differential equation y'' = -3/(t^3). A participant provides an answer involving the expression -3/(2t^-2) - 2t + 7/2 but is questioned about the integration process and the use of variables. Clarifications are sought regarding the correct variable (y or s) and the interpretation of the expression. The conversation emphasizes the need for proper integration and understanding of the terms involved in the equation. Accurate integration is crucial for arriving at the correct solution.
laker_gurl3
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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.

y'=-3\frac{t^{-2}}{-2}+...

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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