Tag Archives: dela Fuente

Elementary Differential Equations by Dela Fuente, Feliciano and Uy Physical Application 2: Exponential Growth and Decay

A certain radioactive material follows the law of exponential change and has a half life of 38 hours. Find how long it takes for 90% of the radioactivity to be dissipated.

Solution:

Use the formula:

S=Ce^{-kt}

First, find the constant of proportionality. In the problem, after 38 hours, half of the radioactivity has been dissipated and a half has been retained. So we can assume that S = 0.5So when t = 38 hrs and C = So.

\left(0.5\right)So=\left(So\right)e^{-k\left(38\right)}

And then solve for k:

k=-0.018241

And then substitute k to the formula:

S=Ce^{-0.018241\left(t\right)}

Now we can solve for the time(t). According to the problem, 90% of the radioactivity is dissipated, so 10% is retained. So we can assume that S = 0.1So and change C = So.

\left(0.1\right)So=\left(So\right)e^{-0.018241\left(t\right)}

And then solve for time(t):

t\:=\:126.23\:hrs
Advertisements
Advertisements

Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 10 Problem 3 — Applications of Ordinary First-Ordered Differential Equations

A tank contains 400 liters of brine. Twelve liters of brine, each containing 2.5 N of dissolved salt, enter the tank per minute, and the mixture, assumed uniform leaves at the rate of 8 liters per min. If the concentration is to be 2 N/litre at the end of one hour, how many newtons of salt should there be present in the tank originally?

Solution:

Consider the following illustration

\frac{dS}{dt}=\left(\frac{dS}{dt}\right)_{en}-\left(\frac{dS}{dt}\right)_{es}

Using,

V_{brine}+\left(rate\:of\:brine\:out\right)t\\\frac{dS}{dt}=\frac{8L}{M}\left(\frac{S}{400+4t}\right)\\=\frac{8S}{\left(400+4t\right)}\\\frac{dS}{dt}=\frac{2S}{\left(100+t\right)}

Using the general solution:

\frac{dS}{dt}=30-\frac{2S}{\left(100+t\right)}\\\frac{dS}{dt}+\frac{2S}{\left(100+t\right)}=30

To solve we will use First Order Linear Differential Equation (FOLDE) where:

P_{\left(t\right)}=\frac{2}{\left(100+t\right)}\:,\:Q_{\left(t\right)}=30

Solve for the integrating factor using the formula:

\sigma =e^{\int \:P_{\left(t\right)}dt}

Apply,

\sigma =e^{\int \:\frac{2}{100+t}dt}\\\sigma =e^{2ln\left(100+t\right)}\\\sigma \:=e^{ln\left(100+t\right)^2}\\\sigma \:=\left(100+t\right)^2

Substitute the given value to the formula:

S\sigma =\int \:\sigma Q\left(t\right)dt+C

Apply,

S\left(100+t\right)^2=\int \:\left(100+t\right)^230dt+C\\S\left(100+t\right)^2=30\int \:\left(100+t\right)^2dt+C\\S\left(100+t\right)^2=30\:\frac{\left(100+t\right)^{^3}}{3}dt+C\\S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C\rightarrow eqn.1

Evaluate C; @t=1hr

Convert 1hr to minutes, where 1hr is simply 60 minutes.

S\left(100+60\right)\:2\:=10\left(100+60\right)\:^3\\C=\frac{2N}{L}\:;\:C=\frac{S}{\left(400+4t\right)}

Get the value of S using the equation:

\:C=\frac{S}{\left(400+4t\right)}

Isolate S,

S=C\left(400+4t\right);\:C=2,\:t=60\\S=2\left(400+4\left(60\right)\right)\\S=1280N

Get the value of C using Eqn.1

S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C; S=1280 , t=60\\
1280\left(100+60\right)^2=10\left(100+60\right)^{^3}+C\\32768000=40960000+C\\32768000-40960000=C\\ C=-8192000

With the presence of the value of C we will now have our working equation:

S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000

Using the given working equation, solve for the value of S @ t=0

S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000; t=0\\S\left(100+0\right)^2=10\left(100+0\right)^{^3}-8192000\\\frac{S\left(1000\right)^2}{1000}=\frac{1808000}{1000}\\S=180.8 N
Advertisements
Advertisements

Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 10 Problem 12 — Applications of Ordinary First-Ordered Differential Equations

A bacterial population follows the law of exponential growth. If between noon and 2 p.m. the population triples, at what time should the population become 100 times what it was at noon? At 10 a.m. what percentage was present?

SOLUTION:

First, we denote

P as the population of bacteria at anytime

Po as the original bacterial population

t = 0 (12 noon)

t = 2 (2 p.m.)

Let us determine the given and the required

GIVEN:

@12nn to 2p.m.; P= 3Po

REQUIRED:

  1. what time should the population become 100 times
  2. at noon
  3. percentage at 10 a.m.

Using the formula of Applications of Ordinary First-Ordered Differential Equations under Exponential Growth or Decay

\frac{dP}{dt}=kP \\
\int \:\frac{dP}{P}=\int \:kdt\\
e^{ln\:P}\:=\:e^{kt\:+\:C}\\
P=\:Ce^{kt}\:\:\:\:\:(Eq.1)\\
@t=0; P=P_o\\
P_o=Ce^{kt}\\
P_o=Ce^{k\left(0\right)}\\
P_o = C

Substituting to Eq.1., we get

P=\:P_{o\:}e^{kt}\:\:\:\:\:\:\:(Eq.2)

Then from the given condition, from 12 noon to 2 p.m., the population triples (using Eq.2), we will solve for the value of k

@t= 2\:;\:P= 3P_o\\
P=\:P_{o\:}e^{kt}\\
3P_{o\:}=\:P_{o\:}e^{k\left(2\right)}\\
k=0.54931

We will then come up with the working equation (WE), this will help us solve the required problems

P_{\:}=\:P_{o\:}e^{\left(0.54931\right)t}

1.) what time should the population become 100 times

Using WE,

t=?\:\:;\:\:P=100P_o\\
P_{\:}=\:P_{o\:}e^{\left(0.54931\right)t}\\
100P_{o\:}=\:P_{o\:}e^{\left(0.54931\right)t}\\
t=8.38\: hrs.\\
t= 8:22:48\: p.m. \; or\:8:23\:p.m.

2.) at noon 

P=P_o

3.) percentage at 10 a.m. 

@10 a.m.\:\:;\:\:t=-2\\
P_{\:}=\:P_{o\:}e^{\left(0.549\right)\left(-2\right)}\\
P_{\:}=\:P_{o\:}\left(0.33333\right)\\
\%=\frac{P}{P_o}{(100)}=\frac{P_o\left(0.33333\right)}{P_o}{(100)}\\
\%=\:33.33\%

Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 2 — Special Second-Ordered Differential Equations

Question:

\frac{d}{dx}\left(x\:\frac{dy}{dx}\:+\left(1+x\right)\:y\right)\:=12\:\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1

Solution:

Since the equation is in the form d/dx (dy/dx + y P(x) = Q(x) , we use Case 1

\begin{align*}
let\:u & = x\frac{dy}{dx}+\left(1+x\right)\:y
\\ \int \:\frac{du}{dx}\:&=\int \:12
\\u &=12x+C_1
\end{align*}

Solving for the value of C1 using the initial values. 

\begin{align*}
x\frac{dy}{dx}+\left(1+x\right)y\: & =12x\:+C_1\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1
\\1(0) + (1+1)0 & = 12(1) + C_1  
\\0+0 & =12+C_1
\\-12 & =C_1
\end{align*}

Rewriting the equation into the general form of a first-order linear differential equation (FOLDE).

\left[x\frac{dy}{dx}+\left(1+x\right)y=12x-12\right]\frac{1}{x}
\\\frac{dy}{dx}+\left(\frac{1+x}{x}\right)y=12-\frac{12}{x}
\\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

Since the equation is now in the form of dy/dx + y P(x) = Q(x), we use FOLDE

\\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

From the general form of a first-order differential equation, we have

\\P \left( x \right)= \left(\frac{1}{x}+1\right) 
\\Q\left( x \right)= 12-\frac{12}{x}

Compute for the integrating factor

\begin{align*}
\phi &= e^ {\int P\left(x \right) dx}
\\\phi & =e^{\int \:\left(\frac{1}{x}+1\right)dx}
\\\phi & \:=e^{\ln x+x}
\\\phi & \:=x\left(e^x\right)
\end{align*}

Substituting everything to the solution of a first-order linear differential equation, we have

y(xe^x)=\int xe^x\left(12-\frac{12}{x}\right)dx+C_2
\\y\left(xe^x\right)=\int \left(12xe^x-12e^x\right)dx+C_2
\\yxe^x=\int 12xe^x-\int \:12e^x\:dx+C_2​

Use Integration by Parts to solve for the first integral

\begin{align*}
\int \:12xe^xdx & = 12 \int xe^x dx\\ 
 u = x &  &du=dx  \\
dv = & e^x \  & v=e^x  \\
\text{Therefore} \\ 
uv-\int \:vdu & =xe^x-\int \:e^xdx \\
& =xe^x-e^x \\
\text{Consequently} \\
\int \:12xe^xdx & = 12 \left( xe^x-e^x\right)
\end{align*}

Therefore,

\begin{align*}
yxe^x & =12\left(xe^x-e^x\right)-12e^x+C_2
\\yxe^x &=12xe^x-12e^x-12e^x+C_2
\\yxe^x &=12xe^x-24e^x+C_2
\end{align*}

Solving for C2

\begin{align*}
yx & =12x-24+\frac{C_2}{e^x}\:\:;\:y=0,\:x=1
\\0\left(1\right) & =12\left(1\right)-24+\frac{C_2}{e^1}\:
\\0 & =12-24e+\frac{C_2}{e^1}\:
\\0 &=-12+\frac{C_2}{e^1}\:
\\12e^1 & =C_2\:
\end{align*}

Therefore, the solution to the problem is

yx=12x-24+\frac{12e^1}{e^x}
\\or
\\yx=12x-24+12e^{1-x}