The rate at which the amount of salt in the tank,
![A(t)](https://tex.z-dn.net/?f=A%28t%29)
, changes over time is given by the ODE
![A'(t)=\dfrac{4\text{ gal}}{1\text{ min}}\cdot\dfrac{1\text{ lb}}{1\text{ gal}}-\dfrac{0.5\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{10+(4-0.5)t\text{ gal}}](https://tex.z-dn.net/?f=A%27%28t%29%3D%5Cdfrac%7B4%5Ctext%7B%20gal%7D%7D%7B1%5Ctext%7B%20min%7D%7D%5Ccdot%5Cdfrac%7B1%5Ctext%7B%20lb%7D%7D%7B1%5Ctext%7B%20gal%7D%7D-%5Cdfrac%7B0.5%5Ctext%7B%20gal%7D%7D%7B1%5Ctext%7B%20min%7D%7D%5Ccdot%5Cdfrac%7BA%28t%29%5Ctext%7B%20lb%7D%7D%7B10%2B%284-0.5%29t%5Ctext%7B%20gal%7D%7D)
and we're told that at the start there is no salt in the tank (fresh water), so
![A(0)=0](https://tex.z-dn.net/?f=A%280%29%3D0)
. The amount of solution in the tank is given by
![10+(4-0.5)t](https://tex.z-dn.net/?f=10%2B%284-0.5%29t)
, so the tank will overflow once this expression is the same as the total tank volume, i.e.
![10+(4-0.5)t=50\implies t=\dfrac{80}7\approx11.43](https://tex.z-dn.net/?f=10%2B%284-0.5%29t%3D50%5Cimplies%20t%3D%5Cdfrac%7B80%7D7%5Capprox11.43)
We find an integrating factor of the form
![\exp\left(\displaystyle\int\frac{\mathrm dt}{20+7t}\right)=\exp\left(\frac17\ln|20+7t|\right)=(20+7t)^{1/7}](https://tex.z-dn.net/?f=%5Cexp%5Cleft%28%5Cdisplaystyle%5Cint%5Cfrac%7B%5Cmathrm%20dt%7D%7B20%2B7t%7D%5Cright%29%3D%5Cexp%5Cleft%28%5Cfrac17%5Cln%7C20%2B7t%7C%5Cright%29%3D%2820%2B7t%29%5E%7B1%2F7%7D)
then distribute this to all terms of the ODE, giving
![A'(t)+\dfrac1{20+7t}A(t)=4](https://tex.z-dn.net/?f=A%27%28t%29%2B%5Cdfrac1%7B20%2B7t%7DA%28t%29%3D4)
![\implies(20+7t)^{1/7}A'(t)+(20+7t)^{-6/7}A(t)=4(20+7t)^{1/7}](https://tex.z-dn.net/?f=%5Cimplies%2820%2B7t%29%5E%7B1%2F7%7DA%27%28t%29%2B%2820%2B7t%29%5E%7B-6%2F7%7DA%28t%29%3D4%2820%2B7t%29%5E%7B1%2F7%7D)
![\left((20+7t)^{1/7}A(t)\right)'=4(20+7t)^{1/7}](https://tex.z-dn.net/?f=%5Cleft%28%2820%2B7t%29%5E%7B1%2F7%7DA%28t%29%5Cright%29%27%3D4%2820%2B7t%29%5E%7B1%2F7%7D)
![(20+7t)^{1/7}A(t)=\dfrac12(20+7t)^{8/7}+C](https://tex.z-dn.net/?f=%2820%2B7t%29%5E%7B1%2F7%7DA%28t%29%3D%5Cdfrac12%2820%2B7t%29%5E%7B8%2F7%7D%2BC)
![A(t)=\dfrac{20+7t}2+\dfrac C{(20+7t)^{1/7}}](https://tex.z-dn.net/?f=A%28t%29%3D%5Cdfrac%7B20%2B7t%7D2%2B%5Cdfrac%20C%7B%2820%2B7t%29%5E%7B1%2F7%7D%7D)
Since
![A(0)=0](https://tex.z-dn.net/?f=A%280%29%3D0)
, we get
![0=\dfrac{20}2+\dfrac C{20^{1/7}}\implies C\approx-15.34](https://tex.z-dn.net/?f=0%3D%5Cdfrac%7B20%7D2%2B%5Cdfrac%20C%7B20%5E%7B1%2F7%7D%7D%5Cimplies%20C%5Capprox-15.34)
so that the amount of salt in the tank is given by
![A(t)=\dfrac{20+7t}2-\dfrac{15.34}{(20+7t)^{1/7}}](https://tex.z-dn.net/?f=A%28t%29%3D%5Cdfrac%7B20%2B7t%7D2-%5Cdfrac%7B15.34%7D%7B%2820%2B7t%29%5E%7B1%2F7%7D%7D)
Then once the tank starts to overflow at
![t\approx11.43](https://tex.z-dn.net/?f=t%5Capprox11.43)
, the amount of salt in the tank is