can you type the answer please.I cant see the hand writing . Thank you Consider

Question

can you type the answer please.I cant see the hand writing . Thank you

Consider the cascade of two tanks shown below with V_1 = 200(gal) and V_2 = 100(gal) being the volumes of brine in the two tanks. Each tank also initially contains 50lbs of salt. The three flow rates indicated in the figure are each 5 gal/min, with pure water flowing into the first. Find a function for the amount, x(t) of salt in tank 1 at time t. Then, with y(t) as the amount of salt in tank 2 at time t. Show that dy/dt = x/40 - y/20 and then solve for your original function.

Explanation / Answer

Given a tank contain with an initial amount v0 of solution . We are pumping in brine with a concentration ci of salt at a rate ri , and such that we are pumping out the solution at a rate of ro. Let   x(t) be the amount of salt in the solution at time t is governed by the differential equation:

dx/dt=rici-(rox/(v+(ri-r0)t)

For the first tank , we have the values ri = ro = 5 gal/min, v = 200 gal, and ci = 0 as pure water is flows into tank 1. Thus the differential equation we obtain is

dx/ dt = 5x(t) /200

By separation of variables we can solve this differential equation we get,

ln|x|=-1/40 t+C1 so x=Ce^(-1/40 t)

Initially we have C = x(0) = 50 and so the final solution is

x(t)=50 e^(-1/40 t) ........(1)

Now let y(t) be the amount of salt in tank 2 then y(t) satisfies the same differential equation above however we now have that the concentration ci of salt in the incoming water changes as a function of time, in fact it is just the total amount of salt x(t) in tank 1 divided by the volume 200 of the solution in tank 1. For tank 2 we check that the other quantities are given by ri = r0 = 5, v = 100, and hence the differential equation we obtain is:

dy/dt=(5x/200 )- (5y/100)=x/40-y/20 ........(2)

=5/4 e^(-1/40 t)- (1/20)y .... from (1)

This is a linear equation dy /dt + (1 /20) y = 5 /4 exp( t /40 ) hence if we multiply both sides of the equation by exp( t 40 and use Leibniz’s rule we have

d/ dx(exp( t /20 )y) = exp( t /20 ) dy/ dx + 1/20 exp( t /20 )y = 5/4 exp(t /20 ).

Integrating we have

exp( t /20 )y = 25 exp(t /20 ) + C.

Initially we have y(0) = 50

therefore 50 = 25+ C hence C = 75, solving for y gives the solution:

y = 25 exp(t /10 ) + 75 exp(t /20 )

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