dy/dt = y/4
dy/y = 1/44 dt
lny = t/4 + c
y = c*e^(t/4)
y(0)=20, so
c*e^0 = 10
c = 20
y = 20e^(t/4)
You are correct.
The particular solution of the differential equation dy/dt equals y over 4 for which y(0) = 20 is
y = 20e−0.25t
y = 19 + e0.25t
y = 20 e0.25t - my answer
y = 20e4t
1 answer