Find the solution to the differential equation...

Find the solution to the differential equation dy/dt = y^2(7 + t), y=8 when t=1. What is y? Any help is appreciated! Thank you!

Separate variables, integrate both sides, then use the fact that y = 8 when t = 1 to solve for the arbitrary constant:

dy/dt = y^2(7 + t)
dy/(y^2) = (7 + t)dt
integral dy/(y^2) = integral (7 + t)dt
-1/y = 7t + (1/2)t^2 + c
y = -1/[7t + (1/2)t^2 + c] = -2/(t^2 + 14t + C) where C = 2c

8 = -2/(1^2 + 14*1 + C) = -2/(15 + C)
15 + C = -2/8 = -1/4
C = -61/4

y = -2/(t^2 + 14t - 61/4)
y = -8/(4t^2 + 56t - 61).

Lord bless you today!

So I worked out the entire problem for ya.
My work is pretty lengthy so I'll just give you the answer.

This is my answer:

y= -1/(7t+1/2t^2-61/8)

Might wanna check that