How do you do these two questions?

T = -22/r + 39

(a) y = [130 − 130e^(-0.015t)] / 3

(b) 22.9

Step-by-step explanation:

d²T/dr² + 2/r dT/dr = 0

If S = dT/dr, then:

dS/dr + 2/r S = 0

dS/dr = -2/r S

dS / S = -2/r dr

ln S = -2 ln r + C

S = e^(-2 ln r + C)

S = Ce^(-2 ln r)

S = C [e^(ln r)]^-2

S = C / r²

dT/dr = C / r²

dT = C dr / r²

T = -C / r + D

When r=1, T=17.

17 = -C + D

When r=2, T=28.

28 = -C/2 + D

Solving:

11 = C/2

C = 22

D = 39

T = -22/r + 39

We can check our answer by finding dT/dr and d²T/dr².

dT/dr = 22/r²

d²T/dr² = -44/r³

Plug into the original equation:

d²T/dr² + 2/r dT/dr = 0

-44/r³ + 2/r (22/r²) = 0

0 = 0

The total flow into the tank is 15 L/min, and the total flow out is 15 L/min, so the volume is constant at 1000 L.  The amount of salt added to the tank is:

(0.05 kg/L) (5 L/min) + (0.04 kg/L) (10 L/min) = 0.65 kg/min

If y is the amount of salt in the tank at time t, and y/1000 is the concentration in the tank at time t, then the amount of salt removed from the tank is:

(y/1000 kg/L) (15 L/min) = 0.015y kg/min

The change in salt over time equals the amount of salt in minus the amount of salt out.

dy/dt = 0.65 − 0.015y

-200 dy/dt = 3y − 130

dy / (3y − 130) = -0.005 dt

3 dy / (3y − 130) = -0.015 dt

ln(3y − 130) = -0.015t + C

3y − 130 = Ce^(-0.015t)

3y = 130 + Ce^(-0.015t)

y = [130 + Ce^(-0.015t)] / 3

At t=0, y=0:

0 = (130 + C) / 3

C = -130

y = [130 − 130e^(-0.015t)] / 3

At t=50:

y = [130 − 130e^(-0.75)] / 3

y ≈ 22.9