Differential Equations of the form dy/dx = g(y)

A balloon is expanding at a rate that is inversely proportional to its volume (V cm^3) at any time t seconds. That is, dV/dt = k/V, where k is a constant.

If the initial volume is 10 cm^3 and after 5 seconds it is 40 cm^3, find:

a) V at any time t.
b) the volume after 8 seconds

I started doing a), here's the working I've done:

dV/dt = k/V

V dV = k dt

∫V dV = ∫k dt

V^2/2 = kt + c

I know this is the right working but I'm not sure where to go from there.

Thank you

a.
Find the general solution by separating the variables then integrating:
dV / dt = k / V
V dV = k dt
∫ V dV = k ∫ 1 dt
V² / 2 = kt + C
V² = 2kt + C
V = √(2kt + C)

Find the particular solution by solving for the constants:
When t = 0, V = 10
√C = 10
C = 100
When t = 5, V = 40
√(10k + C) = 40
10k + C = 1600
10k = 1600 - C
k = 160 - C / 10
k = 160 - 10
k = 150
V = √(300t + 100)

2.
Find the volume by plugging this value of time into the formula:
V(8) = √(2400 + 100)
V(8) = √2500
V(8) = 50 cm³

ok

dV/dt = k/V

V dV = k dt

=> V^2/2 = kt + c

=> V^2(t) = 2kt + C

initialy (t = 0) V = 10 , so :

V^2(0) = C = 10^2 = 100

=> V^2(t) = 2kt + 100

and V(5) = 40

=> 40^2 = 2k * 5 + 100

=> 10k = 1500 => k = 150

hence V^2(t) = 300t + 100

=> V(t) = sqrt (300t + 100) <<<<


b) V(8) = sqrt (300*8 + 100) = sqrt(2500) = 50 cm^3