Tag Archives: Differential Equation

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 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(x1,y1)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−y1=−1m(x−x1)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

−y1=−1m(x−x1)−my1=−1(x−x1)−my1=−x+x1x=x1+my1\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)

(x1−0)2+(y1−0)2=(x1+my1−x1)2+(0−y1)2x12+y12=m2y12+y12x12+y12=m2y12+y12x12=m2y12x1=m1y1    ;m=dydxx1=dydxy1\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,

x=ydydxxdx=ydy\begin{align*} x&=y\frac{dy}{dx}\\ xdx&=ydy \end{align*}

By integrating,

∫ ydx=∫ xdyy22=x22+Cy2=x2+2C\begin{align*} \int \:ydx&=\int \:xdy\\ \frac{y^2}{2}&=\frac{x^2}{2}+C\\ y^2&={x^2}+2C\\ \end{align*}

We get,

y2−x2−=2Cy^2-x^2-=2C
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 1 – Special Second-Ordered Differential Equations

Find the General Solution

ddx(dydx)= 6x + 3\frac{d}{dx}\left(\frac{dy}{dx}\right)=\:6x\:+\:3

Solution:

ddx(dydx)= 6x + 3solve  the  equation  using  case  1,let  u=dydx∫ dudx = ∫ (6x + 3)using  separation  of  variable  divide  both  sides  by  dx,∫ du = ∫ (6x + 3)dxby  integrating  using  the  sum  rule:we  get,u=3x2 + 3x + C1substitute  the  value  of  u=dydxdydx=3x2 + 3x + C1using  separation  of  variable:∫ dy=∫ (3x2+3x + C1)dxapply  the  sum  rule:∫ dy=∫ 3x2dx+∫ 3xdx+∫ C1dxy=x3+3x22+C1x+C2\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\\
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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=0y''+2y'-6x-3=0

Solution:

A second-order differential equation can be written in the form:

ay″+by′+cy=g(x)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+3y''+2y'=6x+3

ii. Let

y′=P=dydxandy′′=dPdxdPdx+P2 = 6x+3\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.

P(x)=2andQ(x)=6x+3dPdx+P2 = 6x+3\begin{align*} P(x) & =2 \\ and \\ Q(x) &=6x+3 \\ \\ \frac{dP}{dx}+P2\:& =\:6x+3 \end{align*}

Find the integrating factor

ɸ=e∫ P(x)dxɸ=e∫  2dxɸ=e2x\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

Pɸ=∫ ɸQ(x)dx+C1Pe2x=∫ e2x(6x+3)dx+C1Pe2x=∫ (e2x6x+3e2x)dx+C1\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

∫ e2x6xdx=6⋅∫ xe2xdx\begin{align*} \int \:e^{2x}6xdx= 6\cdot \int \:xe^{2x}dx \\ \end{align*}

Let u = 2x and du/2 = dx

32∫ euudu\frac{3}{2}\int \:e^uudu

By IBP, Let v=u, dv=du and eudu, n=eu.

nv−∫ndvueu−32∫  eudu=euu−eu=3e2xx−32e2x\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

∫ 3e2xdx\int \:3e^{2x}dx

Let u = 2x and du/2 = dx

32∫ eudu=32eu=32e2x\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

Pe2x=3e2xx−32e2x+32e2x+C1Pe2x=3e2xx+C1\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,

[dydxe2x=3e2xx+C1]1e2xdydx=3x+C1e2x∫ dy=∫ (3x+C1e2x)dx+C2\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=3x22−C1e−2x2+C2y=\frac{3x^2}{2}-\frac{C_1e^{-2x}}{2}+C_2
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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(y)2=0yy''+2\left(y\right)^2=0

Solution:

Based on Special Second-Ordered Differential Equation: Special case 3

F(d2ydx2, dydx, y)=0F\left(\frac{d^2y}{dx^2},\:\frac{dy}{dx},\:y\right)=0

Denote and substitute to the given equation.

P=y′=dydxPdpdy=y′′=d2ydx2P= y' =\frac{dy}{dx} \\ P\frac{dp}{dy}= y'' =\frac{d^2y}{dx^2}

We will have,

y(Pdpdx)+2(P)2=0y(P\frac{dp}{dx})+2(P)^2=0

Divide both sides with

 1yP \:\frac{1}{yP}

We will come to,

dpdy+2Py=0\frac{dp}{dy}+\frac{2P}{y}=0

Tranpose,

2Py\frac{2P}{y}

We will have

dpdy=−2Py\frac{dp}{dy}=-\frac{2P}{y}

Integrate both sides,

∫dpdy=−∫2Py\int \frac{dp}{dy}=-\int\frac{2P}{y}

The equation will become a SEPARABLE DIFFERENTIAL EQUATION, multiply both sides with

dyP \frac{dy}{P}\:

We will come to the equation:

dpP=−2ydy \frac{dp}{P}=-\frac{2}{y}dy

Integrate both sides,

∫dpP=−∫2ydy\int \frac{dp}{P}=-\int\frac{2}{y}dy

The answer will be:

ln⁡(P)=ln⁡(y−2)+lnC\ln \left(P\right)=\ln \left(y^{-2}\right)+lnC

Apply logarithmic definition and exponent rule

logab=c then, b=acab+c=abacloga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c

The answer will be:

P=Cy2P=\frac{C}{y^2}

Recall that

P=dydxP=\frac{dy}{dx}

Substitute the original value of P,

dydx=Cy2\frac{dy}{dx}=\frac{C}{y^2}

Again, this is a Separable Differential Equation, multiply both sides with:

y2dxy^{2}dx

It will become

y2dy=Cdxy^{2}dy=Cdx

Integrate both sides,

∫y2dy=∫Cdx\int y^{2}dy=\int Cdx

The answer will be

y33=C1x+C2\frac{y^3}{3}=C1x+C2

Multiply both sides with 3 and the final answer will be

y3=C1x+C2y^3=C_1x+C_2

You can still solve it explicitly,

y=C1x+C23y=\sqrt[3]{C_1x+C_2}
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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 6 — Special Second-Ordered Differential Equations

Solve the following differential equation

yy′′−(y′)2+y′=0yy''-\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′             Pdpdy=y′′so\:we\:let\:P=y' \\ \:\:\:\:\:\:\:\:\:\:\:\:\:P\frac{dp}{dy}=y''

Substituting to the equation, we get

yPdPdy−P2+P=0yP\frac{dP}{dy}-P^2+P=0

Removing the variables y and P from the 1st term we get

dPdx−Py+1y=0      ,the equation has become a FOLDE  dPdy−Py=−1y           T(y)=−1y,     Q(y)=−1y     ϕ=e∫ −1ydy=y−1         Pϕ=∫ϕ Q(y)dy + C1            Py−1=∫ y−1(−y−1)dy+C1Py−1=∫ −y−2dy+C1 Py=1y+C1    P=1+yC1dydx=1+yC1\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

dy1+yC1= dx∫ dy1+yC1=∫ dx        let u=1+yC1      du=C1dyduC1=dy   1C1∫ duu=∫ dx           1C1ln u=x+C2\\ \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∣1+yC1∣=C1x+C2ln\left|1+yC_{1}\right|=C_{1}x+C_{2}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 1

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PROBLEM:

If f(x)=x2−4x\displaystyle f\left(x\right)=x^2-4x, find

a) f(−5)\displaystyle f\left(-5\right)

b) f(y2+1)\displaystyle f\left(y^2+1\right)

c) f(x+Δx)\displaystyle f\left(x+\Delta x\right)

d) f(x+1)−f(x−1)\displaystyle f\left(x+1\right)-f\left(x-1\right)

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SOLUTION:

Part A

f(−5)=(−5)2−4(−5)=25+20=45  (Answer)\begin{align*} f\left(-5\right) & =\left(-5\right)^2-4\left(-5\right)\\ & =25+20\\ & =45 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

f(y2+1)=(y2+1)2−4(y2+1)=y4+2y2+1−4y2−4=y4−2y2−3  (Answer)\begin{align*} f\left(y^2+1\right) & = \left(y^2+1\right)^2-4\left(y^2+1\right)\\ & =y^4+2y^2+1-4y^2-4\\ & =y^4-2y^2-3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

f(x+Δx)=(x+Δx)2−4(x+Δx)=(x+Δx)[(x+Δx)−4]=(x+Δx)(x+Δx−4)  (Answer)\begin{align*} f\left(x+\Delta x\right)&=\left(x+\Delta x\right)^2-4\left(x+\Delta x\right)\\ & =\left(x+\Delta x\right)\left[\left(x+\Delta x\right)-4\right]\\ & =\left(x+\Delta x\right)\left(x+\Delta x-4\right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part D

f(x+1)−f(x−1)=[(x+1)2−4(x+1)]−[(x−1)2−4(x−1)]=[x2+2x+1−4x−4]−[x2−2x+1−4x+4]=x2−x2+2x−4x+2x+4x+1−4−1−4=4x−8=4(x−2)  (Answer)\begin{align*} f\left(x+1\right)-f\left(x-1\right) & =\left[\left(x+1\right)^2-4\left(x+1\right)\right]-\left[\left(x-1\right)^2-4\left(x-1\right)\right]\\ & = \left[x^2+2x+1-4x-4\right]-\left[x^2-2x+1-4x+4\right]\\ & =x^2-x^2+2x-4x+2x+4x+1-4-1-4\\ & =4x-8\\ & =4\left(x-2\right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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