Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).
Salt flows in at a rate of
and flows out at a rate of
so that the net rate of change in the amount of salt in the tank is given by the linear differential equation
Multiply both sides by the integrating factor, 
, and rewrite the left side as the derivative of a product.
![\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))](https://tex.z-dn.net/?f=%5Cdfrac%20d%7Bdt%7D%5Cleft%5Be%5E%7Bt%2F180%7D%20A%28t%29%5Cright%5D%20%3D%20136%20e%5E%7Bt%2F180%7D%20%281%20%2B%2015%20%5Csin%28t%29%29)
Integrate both sides with respect to t (integrate the right side by parts):
![\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cfrac%20d%7Bdt%7D%5Cleft%5Be%5E%7Bt%2F180%7D%20A%28t%29%5Cright%5D%20%5C%2C%20dt%20%3D%20136%20%5Cint%20e%5E%7Bt%2F180%7D%20%281%20%2B%2015%20%5Csin%28t%29%29%20%5C%2C%20dt)
Solve for A(t) :
The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.
So,
Recall the angle-sum identity for cosine:
so that we can condense the trigonometric terms in A(t). Solve for R and θ :
Recall the Pythagorean identity and definition of tangent,
Then
and
so we can rewrite A(t) as
As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of
and
which is to say, with amplitude
