Answer:
Step-by-step explanation:
From the given information:

Given that the solution is pumped at a slower rate of 4gal/min
Then:


The differential equation can be expressed as:

Integrating the linear differential equation; we have::


multiplying above integrating factor fields; we have:

![\dfrac{d}{dt}\bigg [ (50 +t)^2 A \bigg ] = 3 (50 +t)^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdt%7D%5Cbigg%20%5B%20%2850%20%2Bt%29%5E2%20A%20%5Cbigg%20%5D%20%3D%203%20%2850%20%2Bt%29%5E2)

A = (50 + t) + c(50 + t)²
Using the given conditions:
A(0) = 20
⇒ 20 = 50 + c (50)⁻²
-30 = c(50) ⁻²
c = -30 × 2500
c = -75000
A = (50+t) - 75000(50 + t)⁻²
The no. of pounds of salt in the tank after 35 minutes is:
A(35) = (50 + 35) - 75000(50 + 35)⁻²
A(35) = 85 - 
A(35) =69.6193 pounds
A(35)
70 pounds
Thus; the number of pounds of salt in the tank after 35 minutes is 70 pounds.