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Let $x\left(t\right)$ denote the amount of salt in the tank at time $t$, with $x$ in pounds, and $t$ in minutes.
The rate at which salt enters the tank is 40 gal/min times $1\/100$ lbs/gal, or $2\/5$ lb/min.
The rate at which salt leaves the tank is $\left(\frac{40}{1000}\right)x\=x\/25$ lbs/min.
The rate of change of salt in the tank, namely, $\frac{\mathrm{dx}}{\mathrm{dt}}$ or $\stackrel{\.}{x}$, is then the difference between the rates at which salt enters and leaves the tank. The units would be lbs/min.
Since the tank initially contained 50 lbs of salt, the governing initialvalue problem is
$\stackrel{\.}{x}\=\frac{2}{5}\frac{x}{25}\,x\left(0\right)\=50$
the solution of which is
$x\left(t\right)\=10\+40{\mathit{\ⅇ}}^{t\/25}$
Figure 5.6.5(a) provides a graph of $x\left(t\right)$ for $t\in \left[0\,15\right]$. The salt content in the tank at $t\=15$ is
$x\left(15\right)\=10\+40{e}^{3tsol;5}\doteq 31.95$ lbs


Figure 5.6.5(a) Solution of IVP






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The limiting value of $x\left(t\right)$ is 10 lbs, attained when each of the 1000 gallons of brine in the tank has the concentration of the incoming fluid, namely, $1\/100$ lb/gal. This limiting value is corroborated analytically by letting $t\to \infty$ in $x\left(t\right)\=10\+40{\mathit{\ⅇ}}^{t\/25}$. In fact, this shows $x\>10$ for all finite $t$, an observation useful in a stepwise solution of the IVP by separation of variables.
The differential equation is easily put into the separated form $\frac{\mathrm{dx}}{\frac{2}{5}\frac{x}{25}}\=\frac{25\mathrm{dx}}{10x}equals;\mathrm{dt}$ from which, by antidifferentiation of both sides, it follows that $25\mathrm{ln}lpar;\left10x\rightrpar;equals;tplus;c$. Dividing through by $25$ and exponentiating both sides leads to
$\left10x\right$

$\={e}^{t\/25c\/25}$

$\left10x\right$

$\={e}^{t\/25}{e}^{csol;25}$

$\left10x\right$

$\=A{e}^{tsol;25}$

$10x$

$\=A{e}^{tsol;25}$

$x$

$\=10\+A{e}^{tsol;25}$



In the third line, ${e}^{c\/25}$ is an arbitrary constant $A$ because $c$ is. The absolute values are removed in line four by recalling that $x\>10$ and introducing the appropriate minus sign on the right.
Finally, imposing the initial condition $x\left(0\right)\=50$ gives $50\=10\+A$ so that $A\=40$ and $x\=10\+40{e}^{tsol;25}$.${}$
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