dy/dt = k/y
y dy = k dt
1/2 y^2 = kt + c
y^2 = 2kt + c
y = √(2kt+c)
E does not work, as you can easily see:
y = √(2kt)+4
dy/dt = 2k/(2√(2kt)) = k/√(2kt)
but that is not k/y = k/(√(2kt)+4)
If dy/dt = k/y and k is a nonzero constant, which of the following could be y?
A. y = √(2kt + 16)
B. y = kt + 5
C. y = √(kt + 16)
D. y = 5e^(kt)
E. y = √(2kt) + 4
I got E by finding the integral of y dy = k dt.
1 answer