Category Archives: Mathematics
Algebra ENGG10: Age Word Problem
Peter is 36 years old. Peter is twice as old as Jun was when Peter was as old as Jun is now. How old is Jun?
Solution:
Let x be the age of Jun now.
Consider the following table:
[wpdatatable id=1]
Therefore,
\begin{align*} 36-x & =x-\frac{x}{2} \\ 36-x & = \frac{x}{2} \\ 36 & = \frac{x}{2}+x \\ 36 & = \frac{3x}{2} \\ x & = \frac{36\left( 2 \right)}{3} \\ x & = 24 \ \text{years old} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
Jun is 24 years old now.
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
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
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:
- what time should the population become 100 times
- at noon
- 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 10 Problem 2 — Applications of Ordinary First-Ordered Differential Equations
Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.
Solution:
Plot points on the curve,
A(x_{1},y_{1})
We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.
y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)
As the normal intersects the x-axis, y = 0
Substituting to the previous equation, we get
\begin{align*} -y_{1}&=-\frac{1}{m}\left(x-x_{1}\right)\\ -my_{1}&=-1\left(x-x_{1}\right)\\ -my_{1}&=-x+x_{1}\\ x&=x_{1}+my_{1} \end{align*}
By using distance formula, from the origin (0,0), to point (x1+y1) = from intersection to (x1+y1)
\begin{align*} \sqrt{\left(x_{1}-0\right)^2+\left(y_{1}-0\right)^2}&=\sqrt{\left(x_{1}+my_{1}-x_{1}\right)^2+\left(0-y_{1}\right)^2}\\ \sqrt{x_{1}^2+y_{1}^2}&=\sqrt{m^2y_{1}^2+y_{1}^2}\\ x_{1}^2+y_{1}^2&=m^2y_{1}^2+y_{1}^2\\ x_{1}^2&=m^2y_{1}^2\\ x_{1}&=m_{1}y_{1}\:\:\:\:;m=\frac{dy}{dx}\\ x_{1}&=\frac{dy}{dx}y_{1} \end{align*}
Change x1 and y1 to x and y,
\begin{align*} x&=y\frac{dy}{dx}\\ xdx&=ydy \end{align*}
By integrating,
\begin{align*} \int \:ydx&=\int \:xdy\\ \frac{y^2}{2}&=\frac{x^2}{2}+C\\ y^2&={x^2}+2C\\ \end{align*}
We get,
y^2-x^2-=2C
Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 1 – Special Second-Ordered Differential Equations
Find the General Solution
\frac{d}{dx}\left(\frac{dy}{dx}\right)=\:6x\:+\:3
Solution:
\frac{d}{dx}\left(\frac{dy}{dx}\right)=\:6x\:+\:3 \\ solve\; the\; equation\; using\; case\; 1,\\ let\; u=\frac{dy}{dx} \\ \int \:\frac{du}{dx}\:=\:\int \:\left(6x\:+\:3\right)\\ using\; separation\; of\; variable\; divide\; both\; sides\; by\; dx,\\ \int \:du\:=\:\int \:\left(6x\:+\:3\right)dx\\ by\; integrating\; using\; the\; sum\; rule:\\ we\; get,\\ u=3x^2\:+\:3x\:+\:C_1\\ substitute\; the\; value\; of\; u=\frac{dy}{dx} \\ \frac{dy}{dx}=3x^2\:+\:3x\:+\:C_1\\ using\; separation\; of\; variable:\\ \int \:dy = \int \:\left(3x^2+3x\:+\:C_1\right)dx\\ apply\; the\; sum\; rule:\\ \int \:dy=\int \:3x^2dx+\int \:3xdx+ \int \:C_1dx\\ y=x^3+\frac{3x^2}{2}+C_1x+C_2\\
Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 4 — Special Second-Ordered Differential Equations
Find the general solution of the differential equation
y''+2y'-6x-3=0
Solution:
A second-order differential equation can be written in the form:
ay″ + by' + cy = g(x)
Therefore, the problem given is a second order linear equation.
Here are STEPS on how to get the general solution:
i. simplify the equation to a Second ODE Form
y''+2y'=6x+3
ii. Let
\begin{align*} y' = P=\frac{dy}{dx} \\ \\ and \\ \\ y'' = \frac{dP}{dx} \\ \\ \frac{dP}{dx}+P2\:=\:6x+3 \end{align*}
iii. By recalling, we can see that the equation is in First-order linear differential equation form. Solving the simplified equation using FOLDE.
\begin{align*} P(x) & =2 \\ and \\ Q(x) &=6x+3 \\ \\ \frac{dP}{dx}+P2\:& =\:6x+3 \end{align*}
Find the integrating factor
\begin{align*} ɸ & =e^{\int \:P\left(x\right)dx} \\ ɸ & =e^{\int \:\:2dx} \\ ɸ & =e^{2x} \end{align*}
Substituting the I.F. to the formula
\begin{align*} Pɸ & =\int \:ɸQ\left(x\right)dx+C_1 \\ Pe^{2x} & =\int \:e^{2x}(6x+3)dx+C_1 \\ Pe^{2x} & =\int \:\left(e^{2x}6x+3e^{2x}\right)dx+C_1 \end{align*}
Integrating the first term
\begin{align*} \int \:e^{2x}6xdx= 6\cdot \int \:xe^{2x}dx \\ \end{align*}
Let u = 2x and du/2 = dx
\frac{3}{2}\int \:e^uudu
By IBP, Let v=u, dv=du and eudu, n=eu.
\begin{align*} nv-\int ndv & \\ ue^u-\frac{3}{2}\int \:\:e^udu & = e^uu-e^u \\ & =3e^{2x}x-\frac{3}{2}e^{2x} \end{align*}
for the second term
\int \:3e^{2x}dx
Let u = 2x and du/2 = dx
\begin{align*} \frac{3}{2}\int \:e^udu & =\frac{3}{2}e^u \\ & =\frac{3}{2}e^{2x} \end{align*}
Combining all the solved terms we get
\begin{align*} Pe^{2x} & =3e^{2x}x-\frac{3}{2}e^{2x}+\frac{3}{2}e^{2x}+C_1 \\ Pe^{2x} & =3e^{2x}x+C_1 \end{align*}
Based on the equation that we derived it is now a separable differential equation, therefore,
\begin{align*} &\left[\frac{dy}{dx}e^{2x}=3e^{2x}x+C_1\right]\frac{1}{e^{2x}} \\ \frac{dy}{dx} & =3x+\frac{C_1}{e^{2x}} \\ \int \:dy & =\int \:\left(3x+\frac{C_1}{e^{2x}}\right)dx+C_2 \end{align*}
GENERAL SOLUTION:
y=\frac{3x^2}{2}-\frac{C_1e^{-2x}}{2}+C_2
Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 9 Problem 5 — Special Second-Ordered Differential Equations
Find the general solution of the differential equation
yy''+2\left(y\right)^2=0
Solution:
Based on Special Second-Ordered Differential Equation: Special case 3
F\left(\frac{d^2y}{dx^2},\:\frac{dy}{dx},\:y\right)=0
Denote and substitute to the given equation.
P= y' =\frac{dy}{dx} \\ P\frac{dp}{dy}= y'' =\frac{d^2y}{dx^2}
We will have,
y(P\frac{dp}{dx})+2(P)^2=0
Divide both sides with
\:\frac{1}{yP}
We will come to,
\frac{dp}{dy}+\frac{2P}{y}=0
Tranpose,
\frac{2P}{y}
We will have
\frac{dp}{dy}=-\frac{2P}{y}
Integrate both sides,
\int \frac{dp}{dy}=-\int\frac{2P}{y}
The equation will become a SEPARABLE DIFFERENTIAL EQUATION, multiply both sides with
\frac{dy}{P}\:
We will come to the equation:
\frac{dp}{P}=-\frac{2}{y}dy
Integrate both sides,
\int \frac{dp}{P}=-\int\frac{2}{y}dy
The answer will be:
\ln \left(P\right)=\ln \left(y^{-2}\right)+lnC
Apply logarithmic definition and exponent rule
loga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c
The answer will be:
P=\frac{C}{y^2}
Recall that
P=\frac{dy}{dx}
Substitute the original value of P,
\frac{dy}{dx}=\frac{C}{y^2}
Again, this is a Separable Differential Equation, multiply both sides with:
y^{2}dx
It will become
y^{2}dy=Cdx
Integrate both sides,
\int y^{2}dy=\int Cdx
The answer will be
\frac{y^3}{3}=C1x+C2
Multiply both sides with 3 and the final answer will be
y^3=C_1x+C_2
You can still solve it explicitly,
y=\sqrt[3]{C_1x+C_2}
Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 6 — Special Second-Ordered Differential Equations
Solve the following differential equation
yy''-\left(y'\right)^2+y'=0
Solutions:
Basically, We need to make the orders of each term to 1. To be able to further break down the equation.
so\:we\:let\:P=y' \\ \:\:\:\:\:\:\:\:\:\:\:\:\:P\frac{dp}{dy}=y''
Substituting to the equation, we get
yP\frac{dP}{dy}-P^2+P=0
Removing the variables y and P from the 1st term we get
\frac{dP}{dx}-\frac{P}{y}+\frac{1}{y}=0\:\:\:\:\:\: , the \:equation\:has\:become\:a \:FOLDE\\ \;\\ \frac{dP}{dy}-\frac{P}{y}=-\frac{1}{y}\:\:\:\:\:\:\:\:\:\:\:T\left(y\right)=-\frac{1}{y},\:\:\:\:\:Q\left(y\right)=-\frac{1}{y}\\ \:\\\:\:\:\: \phi =e^{\int \:-\frac{1}{y}dy}\\ =y^{-1} \\\:\:\:\:\:\:\:\:\:P\phi=\int\phi\:Q(y)dy\:+\:C_{1} \\\:\:\:\:\:\:\:\:\:\:\:\:Py^{-1}=\int \:y^{-1}\left(-y^{-1}\right)dy+C_{1} \\ Py^{-1}=\int \:-y^{-2}dy+C_{1} \\ \:\\ \frac{P}{y}=\frac{1}{y}+C_{1} \\\: \:\\\:\: P=1+yC_{1} \\ \frac{dy}{dx}=1+yC_{1}
By means of Separation of Variables
\\ \frac{dy}{1+yC_{1}}=\:dx \\ \int \:\frac{dy}{1+yC_{1}}=\int \:dx \\\:\:\:\:\:\:\:\: let\:u=1+yC_{1} \\\:\:\:\:\:\: du=C_{1}dy\\ \frac{du}{C_{1}}=dy \\\:\:\: \frac{1}{C_{1}}\int \:\frac{du}{u}=\int \:dx \\\:\:\:\:\:\:\:\:\:\:\: \frac{1}{C_{1}}ln\:u=x+C_{2}
We get
ln\left|1+yC_{1}\right|=C_{1}x+C_{2}
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