let me see..,
\[\frac{ dV }{ dt } = 4-2= 2\]
\[\int\limits dV = 2 \int\limits dt\]
\[V = 2t + C\]
because of at t=0 V of water = 50, C = 50, then
\[V = 2t +50\]
\[\frac{ dQ }{ dt }= 0.1 (4) - 2\left( \frac{ Q }{ 2t+50 } \right)\]
\(\frac{ dQ }{ dt } +2\left( \frac{ Q }{ 2t+50 } \right) = 0.1 (4) \) .......(1)
i got p(t) = 2/(2t+50), then for integration factor, i have:
\[\mu(t) = \exp \left( \int\limits_{0}^{t} \frac{ 2 }{ 2t+50 } dt\right) \]
\[\mu(t) = \exp(\ln |2t+50|) = 2t+50\]
then multiplying both side of (1) equations by (2t+50), then:
\[(2t+50)\frac{ dQ }{ dt }+ \frac{ 2 }{ 2t+50 } (2t+50) Q = 0.4 (2t+50)\]
\[\int\limits d \left( (2t+50) Q \right) = 0.4 \int\limits (2t+50) dt\]
\[(2t+50) Q = 0.4(t^{2} + 50t) + C\]
\[Q = \frac{ 0.4t^{2}+ 20t }{ 2t+50 } + C\]
because at t = 0 Q = 0, then i have C = 0, then i have
\(Q = \frac{ 0.4t^{2}+ 20t }{ 2t+50 }\) .......... (2)
(note: it is from t = 0 until t = 25 minutes, and further the volume remains constan at 100 gal),
FOR 25 minutes to 2 hour (120 minutes)
\[\frac{ dQ }{ dt } = 0.1(4) - 4\frac{ Q }{ 100 } \]
\(\frac{ dQ }{ dt } + 4\frac{ Q }{ 100 } = 0.1(4) \) .......(3)
i have p(t), is -4/100, then i have new integration factor:
\[\mu(t) = \exp \left( \int\limits_{0}^{t} \frac{ 4 }{ 100 } dt \right) = \exp \left( \frac{ 4t }{ 100 } \right) = e^{\frac{ 4t }{ 100 }}\]
multiplying both sides of (3) equations by exp (4t/100)
\[e^{\frac{ 4t }{ 100 }}\frac{ dQ }{ dt } + 4 e^{\frac{ 4t }{ 100 }}\frac{ Q }{ 100 } = 0.4 e^{\frac{ 4t }{ 100 }}\]
\[\int\limits d \left( e^{\frac{ 4t }{ 100 }} Q\right) = 0.4 \int\limits \exp \left( \frac{ 4t }{ 100 }\right) dt \]
\[\exp \left( \frac{ 4t }{ 100 } \right) Q = 10 e^{\frac{ 4t }{ 100 }} + C\]
\(Q (t) = 10 + C \exp \left( \frac{ -4t }{ 100 } \right)\) ......(4)
OK.., back to (2) equations.., for t = 25, i have
:
\[Q(25) = \frac{ 0.4(25)^{2} + 20 (25) }{ 2(25) + 50} = \frac{ 0.4(625)+ 500 }{ 100 } = \frac{ 750 }{ 100 } = 7.5\]
So, at t = 25 minutes, Q is equal to 7.5 lb, substitute this to (4)
\[Q (25) = 10 + C \exp \left( \frac{ -4(25) }{ 100 } \right) \]
\[7.5 = 10 + C \exp \left( \frac{- 100 }{ 100 } \right) \]
\[C = -2.5 e\]
finally i got the amount of salt formula as functions of t at the volume remains constant at 100 gal.
\[Q(t) = 10 -(2.5e) \exp \left( \frac{-t }{ 25 } \right) \]
SO, at t = 2 hours (120 minutes), the value of the amount of salt is:
\[Q(120) = 10 -(2.5e) \exp \left( \frac{-(120) }{ 25 } \right) = 10 - 0.05593 \approx 9.944 lb\]