Let [tex]A(t)[/tex] be the amount of fertilizer in the tank at time [tex]t[/tex]. Then
[tex]\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{2\text{ lb}}{1\text{ gal}}\times\dfrac{1\text{ gal}}{1\text{ min}}}_{\text{rate of salt flowing in}}-\underbrace{\dfrac{A(t)\text{ lb}}{100+(1-1)t\text{ gal}}\times\dfrac{1\text{ gal}}{1\text{ min}}}_{\text{rate of salt flowing out}}[/tex]
[tex]\dfrac{\mathrm dA}{\mathrm dt}+\dfrac1{100}A=2[/tex]
This ODE is separable, and you can write
[tex]\dfrac{\mathrm dA}{2-\frac1{100}A}=\mathrm dt[/tex]
Integrating both sides gives
[tex]\displaystyle\int\frac{\mathrm dA}{2-\frac1{100}A}=\int\mathrm dt[/tex]
[tex]\displaystyle100\int\frac{\mathrm dA}{200-A}=\int\mathrm dt[/tex]
[tex]-100\ln|200-A|=t+C[/tex]
[tex]\ln|200-A|=-\dfrac t{100}+C[/tex]
[tex]200-A=Ce^{-t/100}[/tex]
[tex]A=200-Ce^{-t/100}[/tex]
At [tex]t=0[/tex], there is no fertilizer in the tank, so [tex]A(0)=0[/tex]. This means
[tex]0=200-Ce^{-0/100}\implies C=200[/tex]
so that the amount of fertilizer in the tank is given by
[tex]A(t)=200-200e^{-t/100}[/tex]
This function is constantly increasing, so it never achieves its maximum, but there is a strict upper bound as [tex]t\to\infty[/tex]. You have [tex]\lim\limits_{t\to\infty}A(t)=200[/tex] lbs of fertilizer in the tank, which makes sense. Given enough time, the tank will be completely saturated with the fertilizer solution, so 2 lb/gal*100 gal = 200 lbs.
