Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

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

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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.

Step 2: 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" $\begin{array}{rcl} \frac{d y}{d x} & = & \frac{y (lny - lnx + 1)}{x} \\ \frac{d y}{d x} & = & \frac{y (\ln (\frac{y}{x}) + 1)}{x} \\ = & \frac{y}{x} (\ln (\frac{y}{x}) + 1) \\ = & \frac{y}{x} \ln (\frac{y}{x}) + \frac{y}{x} \end{array}$

Step 3: Substitution method

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

Then $y = vx$ .

By Differentiating,

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

Step 4: Integrate the equation

Now, integrate on both sides.

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

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

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

Now,

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

Substitute $w = \ln v$ .

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

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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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: General form of Homogeneous equation

Step 2: Evaluate the given equation

Step 3: Substitution method

Step 4: Integrate the equation

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use the method discussed under “Homogeneous Equations” to solve problems 9-16. dydx=y(lny-lnx+1)x

Step by Step Solution

Step 1: General form of Homogeneous equation

Step 2: Evaluate the given equation

Step 3: Substitution method

Step 4: Integrate the equation

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