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A drug is carried into an organ of volume V cm3cm^3 by a liquid that enters the organ at the rate of a cm3/sec\mathrm{cm}^{3} / \mathrm{sec} and leaves it at the rate of b cm3/sec\mathrm{cm}^{3} / \mathrm{sec}. The concentration of the drug in the liquid entering the organ is c g/cm3\mathrm{g} / \mathrm{cm}^{3}. If the concentration of the drug in the organ at time t is increasing at the rate of x(t)=1V(acbx0)ebtvx^{\prime}(t)=\frac{1}{V}\left(a c-b x_{0}\right) e^{-b t v} g/cm3/sec\mathrm{g} / \mathrm{cm}^{3} / \mathrm{sec} and the concentration of the drug in the organ initially is x0g/cm3x_{0} \mathrm{g} / \mathrm{cm}^{3}, show that the concentration of the drug in the organ at time t is given by x(t)=acb+(x0acb)ebcvx(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 tt is given by

x(t)=acb+(x0acb)ebtV\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

x(t)=1V(acbx0)ebtVx(0)=x0}\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(t)x^\prime(t) we have that

x(t)=1V(acbx0)ebtVdt=1V(acebtVdtbx0ebtVdt)=1V(acebtVdtbx0ebtVdt)\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*}

btV=ubt=Vubdt=Vdudt=Vbdu\Rightarrow\begin{vmatrix} -\frac{bt}{V}&=& u\\ bt&=&-V\cdot u\\ b\, dt&=&-V\, du\\ \, dt&=&-\frac{V}{b}\, du \end{vmatrix}

x(t)=1V[aceu(Vb)dubx0eu(Vb)du]=1V[acVbeudu+bVx0beudu]=acVVbeu+Vbx0bVeu+Cx(t)=acbebtV+x0ebtV+C\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)=x0x(0)=x_0 we obtain

x0=x(0)x0=acbe0+x0e0+Cx0=acb+x0+CC=acb\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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