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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−ktS=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.

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

And then solve for k:

k=−0.018241k=-0.018241

And then substitute k to the formula:

S=Ce−0.018241(t)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.

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

And then solve for time(t):

t = 126.23 hrst\:=\:126.23\:hrs
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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

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

Using,

Vbrine+(rate of brine out)tdSdt=8LM(S400+4t)=8S(400+4t)dSdt=2S(100+t)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:

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

Solve for the integrating factor using the formula:

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

Apply,

σ=e∫ 2100+tdtσ=e2ln(100+t)σ =eln(100+t)2σ =(100+t)2\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:

Sσ=∫ σQ(t)dt+CS\sigma =\int \:\sigma Q\left(t\right)dt+C

Apply,

S(100+t)2=∫ (100+t)230dt+CS(100+t)2=30∫ (100+t)2dt+CS(100+t)2=30 (100+t)33dt+CS(100+t)2=10(100+t)3+C→eqn.1S\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(100+60) 2 =10(100+60) 3C=2NL ; C=S(400+4t)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=S(400+4t)\:C=\frac{S}{\left(400+4t\right)}

Isolate S,

S=C(400+4t); C=2, t=60S=2(400+4(60))S=1280NS=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(100+t)2=10(100+t)3+C;S=1280,t=601280(100+60)2=10(100+60)3+C32768000=40960000+C32768000−40960000=CC=−8192000S\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(100+t)2=10(100+t)3−8192000S\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(100+t)2=10(100+t)3−8192000;t=0S(100+0)2=10(100+0)3−8192000S(1000)21000=18080001000S=180.8NS\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
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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

dPdt=kP∫ dPP=∫ kdteln P = ekt + CP= Cekt     (Eq.1)@t=0;P=PoPo=CektPo=Cek(0)Po=C\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= Po ekt       (Eq.2)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=3PoP= Po ekt3Po = Po ek(2)k=0.54931@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 = Po e(0.54931)tP_{\:}=\:P_{o\:}e^{\left(0.54931\right)t}

1.) what time should the population become 100 times

Using WE,

t=?  ;  P=100PoP = Po e(0.54931)t100Po = Po e(0.54931)tt=8.38 hrs.t=8:22:48 p.m.  or 8:23 p.m.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=PoP=P_o

3.) percentage at 10 a.m.

@10a.m.  ;  t=−2P = Po e(0.549)(−2)P = Po (0.33333)%=PPo(100)=Po(0.33333)Po(100)%= 33.33%@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:

ddx(x dydx +(1+x) y) =12  ; dydx=0, y=0, x=1\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

let u=xdydx+(1+x) y∫ dudx =∫ 12u=12x+C1\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.

xdydx+(1+x)y =12x +C1 ; dydx=0, y=0, x=11(0)+(1+1)0=12(1)+C10+0=12+C1−12=C1\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).

[xdydx+(1+x)y=12x−12]1xdydx+(1+xx)y=12−12xdydx+(1x+1)y=12−12x\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

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

Compute for the integrating factor

ϕ=e∫P(x)dxϕ=e∫ (1x+1)dxϕ =eln⁡x+xϕ =x(ex)\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(xex)=∫xex(12−12x)dx+C2y(xex)=∫(12xex−12ex)dx+C2yxex=∫12xex−∫ 12ex dx+C2​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

∫ 12xexdx=12∫xexdxu=xdu=dxdv=ex v=exThereforeuv−∫ vdu=xex−∫ exdx=xex−exConsequently∫ 12xexdx=12(xex−ex)\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,

yxex=12(xex−ex)−12ex+C2yxex=12xex−12ex−12ex+C2yxex=12xex−24ex+C2\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

yx=12x−24+C2ex  ; y=0, x=10(1)=12(1)−24+C2e1 0=12−24e+C2e1 0=−12+C2e1 12e1=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+12e1exoryx=12x−24+12e1−xyx=12x-24+\frac{12e^1}{e^x} \\or \\yx=12x-24+12e^{1-x}