A tank contains 300 gallons of water and 100 gallons of pollutant. Fresh water is pumped into the tank at rate 2 gal/min, and the well stirred mixture leaves at the same rate. How long does it take for the concentration of pollutants in the tank to decrease to \frac{1}{10} of its original value?
Solution
Let V\left ( t\right ) be the volume in gallons of the pollutant at time t. Hence\begin{equation} \frac{dV\left ( t\right ) }{dt}=R_{in}-R_{out} \tag{1} \end{equation} Where R_{in} is the rate in gallons per min that the pollutant is entering the tank and R_{out} is the rate in gallons per min that the pollutant is leaving the tank. In this problem \begin{equation} R_{in}=0 \tag{1A} \end{equation} Since no pollutant enters the tank. And R_{out}=2 gal/min. But each gallon that leaves contains the ratio \frac{V\left ( t\right ) }{400} of pollutant at any moment of time. This is because the volume of the tank is fixed at 400 gallons since same volume enters as it leaves. Hence \begin{equation} R_{out}=2\frac{V\left ( t\right ) }{400}\qquad \text{gal/min} \tag{1B} \end{equation} Using (1A,1B) in (1) gives\begin{align*} \frac{dV\left ( t\right ) }{dt} & =-\frac{2}{400}V\left ( t\right ) \\ \frac{dV\left ( t\right ) }{dt}+\frac{1}{200}V\left ( t\right ) & =0 \end{align*}
This is a linear ODE. The integration factor is I=e^{\int \frac{1}{200}dt}=e^{\frac{t}{200}}. Therefore the above can be written as \begin{align*} \frac{d}{dt}\left ( V\left ( t\right ) I\right ) & =0\\ \frac{d}{dt}\left ( Ve^{\frac{t}{200}}\right ) & =0 \end{align*}
Integrating gives the general solution as\begin{equation} Ve^{\frac{t}{200}}=C \tag{1} \end{equation} Using initial conditions, at t=0, V=100 gallons. Substituting these in the above to solve for C gives 100=C Hence the solution (1) becomes\begin{equation} V\left ( t\right ) =100e^{\frac{-t}{200}} \tag{2} \end{equation} To find the time t when V\left ( t\right ) =10 gallons (this is \frac{1}{10} of the original volume of pollutant, which is 100 gallons), then the above becomes 10=100e^{\frac{-1}{200}t_{0}} Solving for t_{0} gives\begin{align*} \frac{1}{10} & =e^{\frac{-1}{200}t_{0}}\\ \ln \left ( \frac{1}{10}\right ) & =\frac{-1}{200}t_{0}\\ t_{0} & =-200\ln \left ( \frac{1}{10}\right ) \end{align*}
Hence \fbox{$t_0=460.517$ minutes} This is the time it takes for the pollutant volume to decrease to \frac{1}{10} of its original value in the tank.
Find the orthogonal trajectory of the curve y=c\sin x
Solution
Let \begin{equation} F\left ( x,y,c\right ) =c\sin x-y\tag{1} \end{equation}
Then F_{x}=c\cos x and F_{y}=-1. Hence the slope of the orhogonal projection is given by
\begin{align*} \frac{dy}{dx} & =\frac{F_{y}}{F_{x}}\\ & =\frac{-1}{c\cos x} \end{align*}
From (1), we need to solve for c from F\left ( x,y,c\right ) =0 which gives c\sin x-y=0 or c=\frac{y}{\sin x}. Substituting this back into the above result gives
\begin{align*} \frac{dy}{dx} & =\frac{-1}{\left ( \frac{y}{\sin x}\right ) \cos x}\\ & =\frac{-\sin x}{y\cos x}\\ & =-\frac{1}{y}\tan x \end{align*}
The above gives the ODE to sovle for the orthogonal trajectory curves. This is separable. Integrating gives \int ydy=-\int \tan xdx But \int \tan xdx=-\ln \left \vert \cos \left ( x\right ) \right \vert . Hence the above becomes\begin{align*} \frac{y^{2}}{2} & =\ln \left ( \left \vert \cos \left ( x\right ) \right \vert \right ) +C_{1}\\ y^{2} & =2\ln \left ( \left \vert \cos x\right \vert \right ) +C \end{align*}
Where C=2C_{1}. Solving for y gives two solutions y\left ( x\right ) =\pm \sqrt{2\ln \left ( \left \vert \cos x\right \vert \right ) +C} For illustration, the above was plotted for C=1,2,3,4,5 in the following (shown in red color) against the function \sin \left ( x\right ) (in blue color). It shows the projection curves all cross \sin \left ( x\right ) at 90^{0} everywhere as expected.
The following plot is over a larger x range, from -2\pi to 2\pi
Show that the solution y\left ( t\right ) of the given initial value problem exists on the specified interval. y^{\prime }=y^{2}+\cos \left ( t^{2}\right ) \qquad y\left ( 0\right ) =0;\qquad 0\leq t\leq \frac{1}{2} Solution
Writing the ODE as\begin{align*} y^{\prime } & =f\left ( t,y\right ) \\ & =y^{2}+\cos \left ( t^{2}\right ) \end{align*}
Let R be rectangle 0\leq t\leq \frac{1}{2},y_{0}-b\leq y\leq y_{0}+b. But y_{0}=0 as given. Therefore R=\left [ 0,\frac{1}{2}\right ] \times \left [ -b,b\right ] Now\begin{align*} M & =\max _{\left ( t,y\right ) \in R}\left \vert f\left ( t,y\right ) \right \vert \\ & =\max _{\left ( t,y\right ) \in R}\left \vert y^{2}+\cos \left ( t^{2}\right ) \right \vert \\ & =b^{2}+1 \end{align*}
Hence \alpha =\min \left ( a,\frac{b}{M}\right ) But a=\frac{1}{2},M=b^{2}+1, therefore the above becomes \alpha =\min \left ( \frac{1}{2},\frac{b}{b^{2}+1}\right ) The largest value \alpha can obtain is when g\left ( b\right ) =\frac{b}{b^{2}+1} is maximum. \begin{align*} g^{\prime }\left ( b\right ) & =\frac{\left ( b^{2}+1\right ) -b\left ( 2b\right ) }{\left ( b^{2}+1\right ) ^{2}}\\ & =\frac{b^{2}+1-2b^{2}}{\left ( b^{2}+1\right ) ^{2}}\\ & =\frac{1-b^{2}}{\left ( b^{2}+1\right ) ^{2}} \end{align*}
Hence g^{\prime }\left ( b\right ) =0 gives 1-b^{2}=0 or b=\pm 1. Taking b=1 gives g_{\max }\left ( b\right ) =\frac{1}{1^{2}+1}=\frac{1}{2}. Therefore \begin{align*} \alpha & =\min \left ( \frac{1}{2},\frac{1}{2}\right ) \\ & =\frac{1}{2} \end{align*}
This shows that the solution y\left ( t\right ) exists on t_{0}\leq t\leq t_{0}+\alpha But t_{0}=0,\alpha =\frac{1}{2}, therefore 0\leq t\leq \frac{1}{2} Hence a unique solution exist inside rectangle R=\left [ 0,\frac{1}{2}\right ] \times \left [ -1,1\right ]
Prove that y\left ( t\right ) =-1 is the only solution of the initial value problem y^{\prime }=t\left ( 1+y\right ) \qquad y\left ( 0\right ) =-1 Solution
The solution is found first to show it is y\left ( t\right ) =-1, then using the uniqueness theory, one can show it is unique. The above ODE is separable. Hence\begin{align} \int \frac{dy}{1+y} & =\int tdt\nonumber \\ \ln \left ( \left \vert 1+y\right \vert \right ) & =\frac{t^{2}}{2}+C\nonumber \\ \left \vert 1+y\right \vert & =e^{\frac{t^{2}}{2}+C}\nonumber \\ 1+y & =C_{1}e^{\frac{t^{2}}{2}} \tag{1} \end{align}
Applying initial conditions gives\begin{align*} 1-1 & =C_{1}\\ C_{1} & =0 \end{align*}
Hence the solution (1) becomes\begin{align*} 1+y & =0\\ y\left ( t\right ) & =-1 \end{align*}
To show the above is the only solution we need to show the uniqueness theorem applies to this ODE over all of \Re . Let\begin{align*} y^{\prime } & =f\left ( t,y\right ) \\ & =t\left ( 1+y\right ) \end{align*}
The above shows that f\left ( t,y\right ) is continuous in t over -\infty <t<\infty and continuous in y over -\infty <y<\infty . Now \frac{\partial f}{\partial y}=t Hence \frac{\partial f}{\partial y} is also continuous in y over -\infty <y<\infty . Therefore a solution exist and is unique in any region that includes the initial conditions. Hence the solution y\left ( t\right ) =-1 found above is the only solution.
Using Euler’s method with step size h=0.1, determine an approximate value of the solution at t=1 for y^{\prime }=2ty\qquad y\left ( 0\right ) =2 Which has analytical solution y\left ( t\right ) =2e^{t^{2}}. Compute approximate value at t=1 using just h=0.1, and compare with y(1).
Solution
Euler method is given by\begin{align*} y_{1} & =y_{0}+hf\left ( t_{0},y_{0}\right ) \\ y_{2} & =y_{1}+hf\left ( t_{1},y_{1}\right ) \\ & \vdots \\ y_{k+1} & =y_{k}+hf\left ( t_{k},y_{k}\right ) \end{align*}
Where y_{0}=2 in this problem, and t_{1}=t_{0}+h,t_{2}=t_{1}+h and so on. Where h=0.1. The following table shows the numerical value of y\left ( t\right ) found at each t starting from 0,0.1,0.2,\cdots ,1.0 and comparing it to the exact y\left ( t\right ) and the error at each step using a small Mathematica program which implements the above method.
