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# A drug is carried into an organ of volume V $cm^3$ by a liquid that enters the organ at the rate of a $\mathrm{cm}^{3} / \mathrm{sec}$ and leaves it at the rate of b $\mathrm{cm}^{3} / \mathrm{sec}$. The concentration of the drug in the liquid entering the organ is c $\mathrm{g} / \mathrm{cm}^{3}$. If the concentration of the drug in the organ at time t is increasing at the rate of $x^{\prime}(t)=\frac{1}{V}\left(a c-b x_{0}\right) e^{-b t v}$ $\mathrm{g} / \mathrm{cm}^{3} / \mathrm{sec}$ and the concentration of the drug in the organ initially is $x_{0} \mathrm{g} / \mathrm{cm}^{3}$, show that the concentration of the drug in the organ at time t is given by $x(t)=\frac{a c}{b}+\left(x_{0}-\frac{a c}{b}\right) e^{-b c v}$

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To show that the concentration of the drug in the organ at time $t$ is given by

$\begin{equation} \pmb{x(t)=\frac{ac}{b}+\left(x_0-\frac{ac}{b}\right)e^{-\frac{bt}{V}}} \end{equation}$

we will solve the following initial value problem

$\begin{array}{ccc} x^\prime(t)&=& \dfrac{1}{V}\cdot (ac-bx_0)e^{-\frac{bt}{V}}\\ x(0)&=& x_0 \end{array}\Bigg\}$

Integrating the function $x^\prime(t)$ we have that

\begin{align*} x(t)&=\frac{1}{V}\int (ac-bx_0)e^{-\frac{bt}{V}}\, dt\\ &=\frac{1}{V}\left(\int ace^{-\frac{bt}{V}}\, dt-\int bx_0\cdot e^{-\frac{bt}{V}}\, dt\right)\\ &=\frac{1}{V}\left(ac\int e^{-\frac{bt}{V}}\, dt-bx_0\int e^{-\frac{bt}{V}}\, dt\right) \end{align*}

$\Rightarrow\begin{vmatrix} -\frac{bt}{V}&=& u\\ bt&=&-V\cdot u\\ b\, dt&=&-V\, du\\ \, dt&=&-\frac{V}{b}\, du \end{vmatrix}$

\begin{align*} x(t)&=\frac{1}{V}\left[ac\int e^u\left(-\frac{V}{b}\right)\, du-bx_0\int e^u\left(-\frac{V}{b}\right)\, du\right]\\ &=\frac{1}{V}\left[-\frac{acV}{b}\int e^u\, du+\frac{bVx_0}{b}\int e^u\, du\right]\\ &=-\frac{ac\cancel{V}}{\cancel{V}b}\cdot e^u+\frac{\cancel{Vb}\cdot x_0}{\cancel{bV}}\cdot e^u+C\\\\ & \quad \Rightarrow x(t)=-\frac{ac}{b}\cdot e^{-\frac{bt}{V}}+x_0\cdot e^{-\frac{bt}{V}}+C \end{align*}

Since $x(0)=x_0$ we obtain

\begin{align*} x_0&=x(0)\\ x_0&=-\frac{ac}{b}\cdot e^0+x_0\cdot e^0+C\\ x_0&=-\frac{ac}{b}+x_0+C\Rightarrow\pmb{C=\frac{ac}{b}} \end{align*}

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