A vat holds 100 cubic feet of liquid and is initially full. Liquid which is salt water with 1 gram of salt per 2 cu ft of liquid is being added at a rate of 2 cu ft per minute. It is mixed immediately, and the mixture is draining at 2 cu ft per minute, so the vat stays exactly full. Let y(t) denote the number of grams of salt in the vat at time t. Write the differential equation for y. Use the graphical method to plot two substantially different solutions for y, depending on the initial amount. Explain the difference. Then use the formula for solution of first-order linear DE to write the explicit solution if the vat is initially filled with pure water (i. e., no salt).

explicit solution : y(t) =49 t

Step-by-step explanation:

Given : The vat holds 100 cubic feet of liquid and it is initially full

with salt water that contained 1 gram of salt per 2 cu ft of liquid

attached below is the detailed solution

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

-1, -2, -3, -4, -6

step-by-step explanation: