Answer:
[tex]y=15e^{e^{- \frac{t}{100}}}kg[/tex]
Step-by-step explanation:
Let y(t) represent the amount of salt in the tank after t minutes. Therefore:
[tex]\frac{dy}{dt}=rate\ in-rate\ out[/tex]
Pure water is entering at a rate of 10 L/min, therefore rate in = 0
The tank contains 1000 L of brine (salt water) with 15 kg of dissolved salt, and a mixed solution leaving at 10 L/min hence;
[tex]rate\ out = \frac{y(t)\ kg}{1000\ L}*10\ L/min= \frac{y(t)\ kg}{100\ min}\\\\Therefore:\\\\\frac{dy}{dt}=0- \frac{y(t)}{100} \\\\\frac{dy}{dt}=- \frac{y(t)}{100}\\\\\frac{dy}{y}=- \frac{1}{100}dt\\\\\int\limits{\frac{dy}{y}} =\int\limits- \frac{1}{100}dt\\\\lny=- \frac{t}{100}+C\\\\Taking\ exponetial\ of\ both\ sides:\\\\y=e^{- \frac{t}{100}+C}\\\\y=e^C.e^{- \frac{t}{100}}\\\\y=Ae^{- \frac{t}{100}}(A=e^C)\\\\at\ y=15kg, t=0\\\\15=Ae^{e^{- \frac{0}{100}}}\\\\A=15\\\\[/tex]
[tex]y=15e^{e^{- \frac{t}{100}}}kg[/tex]
