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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=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
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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\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}
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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''-\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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College Physics by Openstax Chapter 3 Problem 13

Find the following for path C in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

Figure 3.56 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

Looking at path C, it moves 1 block upward, 5 blocks to the right, 2 blocks downward, 1 block to the left, 1 block upward, and 3 blocks to the left. So, the total distance is

\begin{align*}
\text{distance} \  = \  &\left( 1\times 120 \ \text{m} \right)+\left( 5\times 120\ \text{m} \right)+\left( 2\times 120\ \text{m} \right)+\left( 1\times 120\ \text{m} \right) \\
& +\left( 1\times 120\ \text{m} \right)+\left( 3\times 120\ \text{m} \right) \\
= \ & 120\ \text{m}+600 \ \text{m}+240 \ \text{m} + 120 \ \text{m}+ 120 \ \text{m}+360\ \text{m} \\
= \ & 1560 \ \text{m} \ \qquad \ {\color{DarkOrange}\left( \text{Answer} \right) }
\end{align*}

Part B

It can be seen from the figure that the end of path C is just one block to the right from the starting point. Therefore, the magnitude of the displacement is 

\begin{align*}
\text{displacement} = 120\ \text{m} \ \qquad \ {\color{DarkOrange}\left( \text{Answer} \right) }
\end{align*}

The direction is to the right or is equivalent to 0° measured from the positive x-axis.

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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}

Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 1 — Special Second-Ordered Differential Equations

Find the general solution of the differential equation

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

Solution:

\begin{align*}
\frac{d}{dx}\left(\frac{dy}{dx}\right) & =6x+3  \\\  \\
let\:u & =\frac{dy}{dx} \\\ \\
\frac{du}{dx} & =6x+3 \\\ \\
Integrate,\\
\int \frac{du}{dx} & =\int (6x+3)dx \\\ \\
\int \frac{du}{dx} & =6\int xdx+3\int dx \\\ \\
u & =\frac{6x^2}{2}+3x+C_1 \\\ \\
u & =3x^2+3x+C_1 \\\ \\
Substitute, \\
\frac{dy}{dx} & =3x^2+3x+C_1 \\\ \\
dy & =\left(3x^2+3x+C_1\right)dx \\\ \\
Integrate,\\
\int dy & =\int (3x^2+3x+C_1)dx \\\ \\
\int dy & =3\int x^2dx+3\int xdx+C_1\int dx \\\ \\
y & =\frac{3x^3}{3}+\frac{3x^2}{2}+C_1x+C_2 \\\ \\
Simplify, \\
y & =x^3+\frac{3x^2}{2}+C_1x+C_2 \\
\end{align*}
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College Physics by Openstax Chapter 3 Problem 7

(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R’=A−B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R”=B−A=−R’ ). Show that this is the case.

Solution:

Refer to this problem.

Part A

Consider Figure 3-7A

Figure 3-7A

First, we solve for the value of α using a simple geometry. 

\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ

We can solve for R using the cosine law.

\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
R & = \sqrt{\left( 12.0\ \text{m} \right)^2+\left( 20.0\ \text{m} \right)^2-2\left(  12.0\ \text{m} \right)\left( 20.0\ \text{m} \right) \cos 110^\circ} \\
R & =  26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Before we can solve for θ, we need to solve for 𝛽 first using the sine law.

\begin{align*}
\frac{\sin \beta}{B} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{B \  \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{B \  \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{20.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 44.9290^\circ
\end{align*}

Now, we can solve for θ. 

\begin{align*}
\theta & = \left( 90^\circ + 20^\circ \right)-\beta \\
\theta & = 110^\circ - 44.9290^\circ \\
\theta & = 65.071 \\
\theta & = 65.1 ^\circ
\end{align*}

Therefore, the compass reading is

65.1^\circ, \text{North of East} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

Part B

Refer to Figure 3-7B

Figure 3-7B

First, we solve for the value of α using a simple geometry. 

\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ

We can solve for R using the cosine law. 

\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
R & = \sqrt{\left( 12.0\ \text{m} \right)^2+\left( 20.0\ \text{m} \right)^2-2\left(  12.0\ \text{m} \right)\left( 20.0\ \text{m} \right) \cos 110^\circ} \\
R & =  26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Before we can solve for θ, we need to solve for 𝛽 first using the sine law. 

\begin{align*}
\frac{\sin \beta}{A} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{A \  \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{A \  \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{12.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 25.0708^\circ
\end{align*}

Now, we can solve for θ. 

\begin{align*}
\theta & = 40^\circ + \beta \\
\theta & = 40^\circ +  25.0708^\circ\\
\theta & = 65.0708^\circ\\
\theta & = 65.1 ^\circ
\end{align*}

Therefore, the compass reading is 

65.1^\circ, \text{South of West} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

This is consistent with Part A because (A-B) = -(B-A).

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College Physics by Openstax Chapter 3 Problem 6

Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B , which is 20.0 m in a direction exactly 40º south of west, and then leg A , which is 12.0 m in a direction exactly 20º west of north. (This problem shows that A+B=B+A.)

Solution:

Consider Figure 3-6A below with B drawn first before A.

Figure 3-6A

Compute for the value of angle β by adding 20° and the complement of 40°. This is by simple geometry. 

\begin{align*}
\beta & = 20^\circ +\left( 90^\circ -40^\circ \right) \\
\beta & = 70^\circ \\
\end{align*}

Solve for the magnitude of R using cosine law. 

\begin{align*}
R^2 & = A^2 +B^2 - 2 A B \cos \beta \\
R & = \sqrt{A^2 +B^2 - 2 A B \cos \beta} \\
R & = \sqrt{\left( 12.0 \ \text{m} \right)^2+\left( 20.0 \ \text{m} \right)^2-2\left( 12.0 \ \text{m} \right)\left( 20.0 \ \text{m} \right) \cos 70^\circ} \\
R & = 19.4892 \ \text{m}\\
R & = 19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

To solve for θ, we need to solve angle α first. This can be done using the sine law.

\begin{align*}
\frac{\sin \alpha}{A}  & = \frac{\sin \beta }{R} \\
\frac{\sin \alpha}{12.0 \ \text{m}}  & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\
\sin \alpha & = \frac{12.0 \ \text{m}\ \sin 70^\circ}{19.4892 \ \text{m}} \\
\alpha & = \sin ^{-1}  \left(  \frac{12.0 \ \text{m}\ \sin 70^\circ}{19.4892 \ \text{m}} \right) \\
\alpha & = 35.3516 ^ \circ
\end{align*}

Finally, we can solve for θ.

\begin{align*}
\theta &  = 40^\circ - 35.3516^\circ  \\
& = 4.6484 ^\circ \\
& = 4.65 ^\circ
\end{align*}

Therefore, the compass reading is

4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}

This is the same with Problem 5.

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College Physics by Openstax Chapter 3 Problem 5

Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.54, then this problem finds their sum R=A+B.)

Figure 3.54

Solution:

Consider Figure 3.5A shown below.

Figure 3.5A

Before we can use cosine law to solve for the magnitude of R, we need to solve for the interior angle 𝛽 first. The value of 𝛽 can be calculated by inspecting the figure and use simple knowledge on geometry. It is equal to the sum of 20° and the complement of 40°. That is

\beta = 20^\circ +\left( 90^\circ -40^\circ  \right) = 70^\circ

We can use cosine law to solve for R.

\begin{align*}
R^2 & =A^2+B^2 -2AB \cos \beta \\
R^2 & = \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ \\
R & = \sqrt{ \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ} \\
R  & =19.4892 \ \text{m} \\
R & =19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

We can solve for α using sine law.

\begin{align*}
\frac{\sin \alpha}{B} & = \frac{\sin \beta}{R} \\
\frac{\sin \alpha}{20.0\ \text{m}} & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\
\sin \alpha & = \frac{20.0 \ \sin 70^\circ }{19.4892} \\
\alpha & = \sin ^{-1}  \left(  \frac{20.0 \ \sin 70^\circ }{19.4892}  \right) \\
\alpha & = 74.6488 ^\circ 
\end{align*}

Then we solve for the value of θ by subtracting 70° from α.

\theta=74.6488 ^\circ -70 ^\circ = 4.65^\circ

Therefore, the compass reading is

4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}
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College Physics by Openstax Chapter 3 Problem 4

Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements \vec{A} and \vec{B} , as in Figure 3.53, then this problem asks you to find their sum \vec{R}=\vec{A}+\vec{B} .)

Figure 3.53

Solution:

Figure 3.4A

Consider Figure 3.54A.

The resultant of the two vectors \vec{A} and \vec{B} is labeled \vec{R}. This \vec{R} is directed \theta ^{\circ} from the x-axis.

We shall use the right triangle formed to solve for the unknowns.

Solve for the magnitude of the resultant.

\begin{align*}
R & = \sqrt{A^2 +B^2} \\
R & = \sqrt{\left(18.0 \ \text{m}  \right)^2+\left( 25.0 \ \text{m} \right)^2} \\
R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Solve for the value of \theta .

\begin{align*}
\theta & = \arctan \left( \frac{B}{A} \right) \\
\theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\
\theta & = 54.2^\circ 
\end{align*}

We need the complementary angle for the compass angle. 

\begin{align*}
90^\circ -54.2^\circ =35.8^\circ 
\end{align*}

Therefore, the compass angle reading is

\begin{align*}
35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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