Water with a small salt content (5 lb in10gal) is flowing into a very salty lake at the rate of$\mathbf{4}\mathbf{·}{\mathbf{10}}^{\mathbf{5}}$gal per hr. The salty water is flowing out at the rate of${\mathbf{10}}^{\mathbf{5}}$gal per hr. If at some time (say t=0)the volume of the lake is${\mathbf{10}}^{\mathbf{9}}$gal, and its salt content is${\mathbf{10}}^{\mathbf{7}}$lb, find the salt content at a time t. Assume that the salt is mixed uniformly with the water in the lake at all times.

The linear differential equation is defined by$\mathbf{\text{x'}}\mathbf{+}\mathbf{\text{Px}}\mathbf{=}\mathbf{\text{Q}}$, whereP and Qare numeric constants or functions in x. It is made up of a yand a yderivative. The differential equation is called the first-order linear differential equation because it is a first-order differentiation.

Given that a lake has a volume of${10}^9$ gal and the lake receive the water at a rate of $4·{10}^4$gal/hr and loses water with a rate of ${10}^5$gal/hr. The water in the lake receives a salt content of $5·{10}^{-3}$IB/gal

The rate of change of salt content of the lake will be:

$\frac{dS}{dt}=\frac{d{S}_{\text{in}}}{dV}\frac{d{V}_{\text{in}}}{dt}-\frac{d{S}_{\text{out}}}{dV}\frac{d{V}_{\text{out}}}{dt}$

Where

$$\begin{gathered} \frac{{\text{dS}}_{\text{in}}}{\text{dV}}=\text{5}\times {\text{10}}^{-\text{3}}\text{Ib/gal,} \\ \frac{{\text{dV}}_{\text{in}}}{\text{dt}}=\text{4}\times {\text{10}}^{\text{5}}\text{gal/hr,} \\ \frac{{\text{dS}}_{\text{out}}}{\text{dV}}=\frac{\text{S}}{{\text{10}}^{\text{9}}+\text{3}\times {\text{10}}^{\text{5}}\text{t}}\text{Ib/gal,} \\ \frac{{\text{dV}}_{\text{in}}}{\text{dt}}=\text{1}\times {\text{10}}^{\text{5}}\text{gal/hr} \end{gathered}$$

The equation in$\text{x'}+\text{Px}=\text{Q}$ form will be:

$$\begin{gathered} {\text{S}}^{\text{'}}=\text{2}\times {\text{10}}^{\text{3}}-\frac{\text{S}}{{\text{10}}^{\text{4}}\text{+ 3t}} \\ {\text{S}}^{\text{'}}+\frac{\text{S}}{{\text{10}}^{\text{4}}+\text{3t}}=\text{2}\times {\text{10}}^{\text{3}} \end{gathered}$$

From the equation 3.4

$$\begin{gathered} \text{I}=\int \frac{\text{dt}}{{\text{10}}^{\text{4}}+\text{3t}} \\ =\frac{\text{1}}{\text{3}}\text{ln}\left({\text{10}}^{\text{4}}+\text{3t}\right) \\ {\text{e}}^{\text{I}}={\text{e}}^{\text{ln}{\left({\text{10}}^{\text{4}}+\text{3t}\right)}^{\text{1/3}}} \\ ={\left({\text{10}}^{\text{4}}+\text{3t}\right)}^{\text{1/3}} \end{gathered}$$

From equation 3.9,

$$\begin{gathered} {\text{Se}}^{\text{'}}\text{=∫}\left(2\times {10}^3\right){\left({10}^4+3t\right)}^{1/3}\text{dt} \\ =\left(\frac{1}{2}\times {10}^3\right){\left({10}^4+3t\right)}^{4/3}dt+S_{\circ } \\ S=\left(\frac{1}{2}\times {10}^3\right)\left({10}^4+3t\right)\frac{S_{\circ }}{{\displaystyle {\left({10}^4+3t\right)}^{1/3}}} \end{gathered}$$

Using the initial equation${\text{S}}_{\text{o}}={\text{10}}^{\text{7}}$, the integration constant${\text{S}}_{\text{o}}$ will be:

$$\begin{gathered} {\text{10}}^{\text{7}}=\left(\frac{\text{1}}{\text{2}}\times {\text{10}}^{\text{7}}\right)+\frac{{\text{S}}_o}{{\text{10}}^{\text{4/3}}} \\ {\text{S}}_o={\text{10}}^{\text{4/3}}\left(\frac{\text{1}}{\text{2}}\times {\text{10}}^{\text{7}}\right) \end{gathered}$$

Then the salt content at any time will be:

$$\begin{gathered} \text{S(t)}=\frac{\text{1}}{\text{2}}\times {\text{10}}^{\text{3}}\left[\left({\text{10}}^{\text{4}}+\text{3t}\right)+\frac{{\text{10}}^{\text{4/3}}{\text{10}}^{\text{4}}}{{\left({\text{10}}^{\text{4}}+\text{3t}\right)}^{\text{1/3}}}\right] \\ =\frac{\text{1}}{\text{2}}\times {\text{10}}^{\text{7}}\left[\left(\text{1}+\frac{\text{3t}}{{\text{10}}^{\text{4}}}\right)+{\left(\text{1}+\frac{\text{3t}}{{\text{10}}^{\text{4}}}\right)}^{\text{- 1/3}}\right] \end{gathered}$$

So, the salt content at any time will be:

$\text{S(t)}=\frac{\text{1}}{\text{2}}\times {\text{10}}^{\text{7}}\left[\left(\text{1}+\frac{\text{3t}}{{\text{10}}^{\text{4}}}\right)+{\left(\text{1}+\frac{\text{3t}}{{\text{10}}^{\text{4}}}\right)}^{\text{- 1/3}}\right]$