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August 30, 2014

August 30, 2014

Posted by **Carly** on Friday, July 29, 2011 at 12:25am.

and find the particular solution satisfying the initial condition y(0)=-3

y(t)=?

- Calculus -
**drwls**, Friday, July 29, 2011 at 6:58amIntegral of dt = Integral of (1/4)y^-6 dy

t = -1/(20 y^-5) + C

0 = 1/20*243 +C

t = -1/(20 y^-5) -1/4860

1/(20 y^-5) = -(1/4860) - t

y^5 = (1/20)/[(-1/4860) - t]

= 243/[-1 -4860 t]

y = -3*[1 +4860 t]^(1/5)

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