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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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$Header Photos for Elementary Differential Equations Chapter 9 Special Second Ordered Equations$

Chapter 9: Special Second-Ordered Equations

Supplementary Problems

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

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Determining if a given Differential Equation is Separable or Not

Determine whether each of the following differential equations is or is not separable, and, if it is separable, rewrite the equation in the form dy/dx=f(x) g(y).
$\qquad \textbf{a}) \quad \frac{dy}{dx}=xy-3x-2y+6$
$\qquad \textbf{b})\quad \frac{dy}{dx}=\sin \left( x+y \right)$
$\qquad \textbf{c}) \quad y\frac{dy}{dx}=e^{x-3y^2}$

Solution:

Part A

\begin{align*} \frac{dy}{dx} & = xy-3x-2y+6 \\ \frac{dy}{dx} & = \left( xy-3x \right)-\left( 2y-6 \right)\\ \frac{dy}{dx} & = x\left( y-3 \right)-2\left( y-3 \right)\\ \frac{dy}{dx} & = \left( x-2 \right)\left( y-3 \right) \end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.

Part B

\begin{align*} \frac{dy}{dx} & = \sin\left( x+y \right) \\ \frac{dy}{dx} & = \sin\left( x \right)\cos\left( y \right) +\cos\left( x \right)\sin\left( y \right)\\ \end{align*}

Since F(x,y) is not factorable in the form f(x) g(y), the given differential equation is not separable.

Part C

\begin{align*} y \frac{dy}{dx} & = e^{x-3y^2}\\ y \frac{dy}{dx} & = \frac{e^x}{e^{3y^2}}\\ \frac{dy}{dx} & =\frac{e^x}{y e^{3y^2}} \\ \frac{dy}{dx} & = e^x \left( \frac{1}{ye^{3y^2}} \right) \\ \end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.

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Mean and Variance | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.10

For the data of Exercise 1.4 on page 13, compute both the mean and variance in “flexibility” for both company A and company B.

Solution:

For the company A:

We know, based on our answer in Exercise 1.4, that the sample mean for samples in Company A is $\displaystyle 7.950$ .

To compute for the sample variance, we shall use the formula

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$ , so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.