A tank with capacity of 600 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of

Step-by-step explanation:

From the information given:

Tank's Capacity = 600 gal

Original content of water = 200 gal

Salt solution = 100 lb

In the tank:

Suppose x(t) = amount of salt at time (t); &

V(t) = volume of water

Then,

V(t) = 200 + t

and

From the above linear equation, the integrating factor can be computed as:

= (200 + t)²

The general solution can now be expressed as:

We know that C = integrating factor, thus taking the integral:

At the initial condition, x(0) = 100

∴

C = ((200)²×100) - 200³

C = -4 × 10⁶

Hence, at any time t, the amount of salt is: