Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 2: Q16E (page 76)

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16. $\frac{d y}{d x} = \frac{y (lny - lnx + 1)}{x}$

Short Answer

Expert verified

Homogeneous equation for the given equation is $y = {xe}^{Cx}$ .

Step by step solution

General form of Homogeneous equation

If the right-hand side of the equation $\frac{dy}{dx} = f (x, y)$ can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

Evaluate the given equation

Given, $\frac{d y}{d x} = \frac{y (lny - lnx + 1)}{x}$ .

Evaluate it.

Since, $\ln (\frac{M}{N}) = \ln M - \ln N$

localid="1655201008520" $\begin{array}{rcl} \frac{d y}{d x} & = & \frac{y (lny - lnx + 1)}{x} \\ \frac{d y}{d x} & = & \frac{y (\ln (\frac{y}{x}) + 1)}{x} \\ = & \frac{y}{x} (\ln (\frac{y}{x}) + 1) \\ = & \frac{y}{x} \ln (\frac{y}{x}) + \frac{y}{x} \end{array}$

Substitution method

Let us take $v = \frac{y}{x}$ .

Then $y = vx$ .

By Differentiating,

$\begin{array}{rcl} \frac{d y}{d x} & = & v + x \frac{d v}{d x} \\ vln v + v & = & v + x \frac{d v}{d x} \\ vln v & = & x \frac{d v}{d x} \\ \frac{1}{vln v} dv & = & \frac{1}{x} dx \end{array}$

Integrate the equation

Now, integrate on both sides.

$\begin{array}{rcl} 鈭� \frac{1}{vln v} dv & = & 鈭� \frac{1}{x} dx \\ 鈭� \frac{1}{vln v} dv & = & \ln |x| + C_{1} \end{array}$

Integrate $\begin{array}{rcl} 鈭� \frac{1}{vln v} dv \end{array}$ separately.

Let us take $w = \ln v$ . Then, $dv = v dw$

Now,

$\begin{array}{rcl} 鈭� \frac{1}{vw} vdw & = & 鈭� \frac{1}{w} dw \\ = & \ln |w| \end{array}$

Substitute $w = \ln v$ .

$\begin{array}{rcl} 鈭� \frac{1}{v \ln v} d v & = & \ln |w| \\ = & \ln |\ln v| \end{array}$

Then,

$\begin{array}{rcl} \ln |\ln v| & = & \ln |x| + C_{1} \\ \ln v & = & e^{\ln |x| + C_{1}} \\ \ln v & = & {xe}^{C_{1}} \\ \ln v & = & {xC}_{2} \\ v & = & e^{{xC}_{2}} \\ v & = & e^{xC} \end{array}$

Substitute $v = \frac{y}{x}$

localid="1655200766733" $\begin{array}{rcl} \frac{y}{x} & = & e^{Cx} \\ y & = & {xe}^{Cx} \end{array}$

Therefore, Homogeneous equation for the given equation is $y = {xe}^{Cx}$ .

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91影视

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16. $\frac{d y}{d x} = \frac{y (lny - lnx + 1)}{x}$

Short Answer

Step by step solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

Integrate the equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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91影视

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16. dydx=y(lny-lnx+1)x

Short Answer

Step by step solution

General form of Homogeneous equation

Evaluate the given equation

Substitution method

Integrate the equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Geometry

Mechanics Maths

Pure Maths

Discrete Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

Use the method discussed under 鈥淗omogeneous Equations鈥� to solve problems 9-16. $\frac{d y}{d x} = \frac{y (lny - lnx + 1)}{x}$