Calculus help

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
 

dextercioby

Science Advisor
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Not really.You should have integrated twice correctly.

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

Daniel.
 

HallsofIvy

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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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