Answer:
The amounts of salt in the tank at any time prior to the instant when the solution begins to overflow is.
![Q(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t](https://tex.z-dn.net/?f=Q%28t%29%3D-%5Cfrac%20%7B4%5Ctimes%2010%5E6%7D%7B%28200%2Bt%29%5E2%7D%2B200%2Bt)
Step-by-step explanation:
Let Q(t) be the amount of salt in the tank at any time
.
Then, its time of change,
by (Balance law).
Since three gallons of salt water runs in the tank per minute, containing
lb of salt, the salt inflow rate is
![3\cdot 1=3](https://tex.z-dn.net/?f=3%5Ccdot%201%3D3)
The amount of water in the tank at any time
is,
![V(t)=200+(3-2)t=200+t,](https://tex.z-dn.net/?f=V%28t%29%3D200%2B%283-2%29t%3D200%2Bt%2C)
Now, the outflow is
gal of the solution in a minute. That is
of the total solution content in the tank, hence
of the salt salt content
, that is
,
Initially, the tank contains
lb of salt,
Therefore, we obtain the initial condition ![Q(0)=100](https://tex.z-dn.net/?f=Q%280%29%3D100)
Thus, the model is
![Q'(t)=3-\frac {2Q(t)}{200+t}, \;\;Q(0)=100](https://tex.z-dn.net/?f=Q%27%28t%29%3D3-%5Cfrac%20%7B2Q%28t%29%7D%7B200%2Bt%7D%2C%20%5C%3B%5C%3BQ%280%29%3D100)
![\Rightarrow Q'(t)+\frac {2}{200+t}Q(t)=3, \;\;Q(0)=100](https://tex.z-dn.net/?f=%5CRightarrow%20Q%27%28t%29%2B%5Cfrac%20%7B2%7D%7B200%2Bt%7DQ%28t%29%3D3%2C%20%5C%3B%5C%3BQ%280%29%3D100)
here,
Linear ODE
So, the integrating factor is
![e^{\int pdt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2](https://tex.z-dn.net/?f=e%5E%7B%5Cint%20pdt%7D%3De%5E%7B2%5Cint%20%5Cfrac%7Bdt%7D%7B200%2Bt%7D%3De%5E%7B%5Cln%28200%2Bt%29%5E2%7D%3D%28200%2Bt%29%5E2)
and the general solution is
![Q(t)(200+t)^2=\int q(t)(200+t)^2dt+c](https://tex.z-dn.net/?f=Q%28t%29%28200%2Bt%29%5E2%3D%5Cint%20q%28t%29%28200%2Bt%29%5E2dt%2Bc)
![\Rightarrow Q(t)=\frac 1{(200+t)^2}\int 3(200+t)^2dt+c](https://tex.z-dn.net/?f=%5CRightarrow%20Q%28t%29%3D%5Cfrac%201%7B%28200%2Bt%29%5E2%7D%5Cint%203%28200%2Bt%29%5E2dt%2Bc)
![=\frac c{(200+t)^2}+200+t](https://tex.z-dn.net/?f=%3D%5Cfrac%20c%7B%28200%2Bt%29%5E2%7D%2B200%2Bt)
using initial condition and find the value of constant c.
![100=Q(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}](https://tex.z-dn.net/?f=100%3DQ%280%29%3D%5Cfrac%20c%7B%28200%2B0%29%5E2%7D%2B200%2B0%5CRightarrow%20-100%3D%5Cfrac%20c%7B200%5E2%7D)
![\Rightarrow c=-4000000=-4\times 10^6](https://tex.z-dn.net/?f=%5CRightarrow%20c%3D-4000000%3D-4%5Ctimes%2010%5E6)
![\Rightarrow Q(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t](https://tex.z-dn.net/?f=%5CRightarrow%20Q%28t%29%3D-%5Cfrac%20%7B4%5Ctimes%2010%5E6%7D%7B%28200%2Bt%29%5E2%7D%2B200%2Bt)
Hence, is the amount of salt in the tank at any moment t.