Q1.

That looks good to me, but I'm very confused by what you then did.

It's a linear differential equation requiring an 'integrating factor'.

Re-write as

The integrating factor is just (t + 25)

t= 0 A = 0 makes C = 0

Put t = 25

So concentration is

Bob

]]>For 2, you should first find the constant c from the two initial conditions given, and only then find the flow rate.

]]>Problem1:

A tank initially holds 25 gal of water. Salt enters at a rate of 2gal/min and the mixture leaves at a rate of 1 gal/min. What will be the concentration of salt when 50 gal of fluid is in the tank?

dA/dt=2-[A(t)/(t+25)]

IF=(t+25)

A(t+25)=(t^2)+50t+c

A=[(t^2)+50t+c]/(t+25)

For A(0)=0; c=0

A=[(t^2)+50t]/(t+25)

50=[(t^2)+50t]/(t+25)

t=35.36

35.36/50=.71

The answer should be 75%.

Problem 2:

A tank initially contains 100 gal of a solution that holds 30lb of a chemical. A solution containing 2 lb/gal of the chemical runs in at a constant rate and the mixture runs out at the same rate. What should be the rate of flow if the tank is to contain 70lb of the chemical after 40min?

dA/dt=2r-(A(t)*r/100)

A(t)=ce^(-rt/100)+200

70=ce^(-40r/100)+200

-130=ce^(-2r/5)

ln(-130)=ln(c)*ln(e^-2r/5)

ln(-130)=c*2r/5

(5/2)ln(-130)=r

The answer should be (5/2)ln(17/13)=.671

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