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

Question: In Problems 33–40, solve the equation given in:
Problem 8.

The solution of the given equation in problem 8 is ${(\frac{y + θ}{y})}^{2} = θ C$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept and definition

When the equation is of the form, $\frac{d y}{d x} = G (\frac{y}{x})$ , then the substitution $v = \frac{y}{x}$ transforms the given equation into a separable equation in the variables v and x.

Step 2: Analyzing the given statement

One has to solve the equation given in problem 7, i.e.,

$(y^{3} - θ y^{2}) d θ + 2 θ^{2} y d y = 0 ........ (1)$

Step 3: Rewriting the equation (1) in the form dydx=G(yx)

Rewriting the equation (1) as,

$\begin{array}{l} \frac{d y}{d θ} = - \frac{(y^{2} - θ y)}{2 θ^{2}} \\ \frac{d y}{d θ} = - \frac{1}{2} (\frac{y^{2}}{θ^{2}} - \frac{y}{θ}) ...... (2) \end{array}$

Step 4: Substituting t=yθ in equation (2) to transform the equation in the variables t and θ

Substitute $t = \frac{y}{θ}$ in equation (2),

So, $d t = \frac{θ d y - y d θ}{θ^{2}}$

$θ^{2} d t = θ d y - y d θ$

Dividing both sides by $d θ$ ,

$\begin{matrix} θ^{2} \frac{d t}{d θ} = θ \frac{d y}{d θ} - y \\ θ^{2} \frac{d t}{d θ} + y = θ \frac{d y}{d θ} \\ \frac{d y}{d θ} = θ \frac{d t}{d θ} + \frac{y}{θ} \\ \frac{d y}{d θ} = θ \frac{d t}{d θ} + t \end{matrix}$

Therefore, equation (2) becomes,

$\begin{matrix} θ \frac{d t}{d θ} + t = - \frac{1}{2} (t^{2} - t) \\ 2 θ \frac{d t}{d θ} + 2 t = - t^{2} + t \\ 2 θ \frac{d t}{d θ} = - t^{2} + t - 2 t \\ 2 θ \frac{d t}{d θ} = - t^{2} - t \end{matrix}$

$\frac{d t}{d θ} = \frac{- t (t + 1)}{2 θ} ...... (3)$

Step 5: Find the solution of equation (3) dtdθ=-t(t+1)2θ.

Separate the variables in equation (3),

$\frac{- 2}{t (t + 1)} d t = \frac{1}{θ} d θ$

Integrating both sides,

$\int \frac{- 2}{t (t + 1)} d t = \int \frac{1}{θ} d θ \cdot \cdot \cdot \cdot \cdot \cdot (4)$

The left hand side of the equation (4) can be solved by using partial fraction,

Therefore,

$\frac{- 2}{t (t + 1)} = \frac{A}{t} + \frac{B}{t + 1}$

Multiply both sides by $t (t + 1)$ ,

$\begin{matrix} - 2 = A (t + 1) + B t \cdot \cdot \cdot \cdot \cdot \cdot (5) \\ A = - 2 \end{matrix}$

Put t = -1 in equation (5),

$\begin{matrix} - 2 = - B \\ B = 2 \end{matrix}$

Therefore, equation (4) becomes,

$\int \frac{- 2}{t} d t + \int \frac{2}{t + 1} d t = \int \frac{1}{θ} d θ$

Put t = 0 in equation (5),

$- 2 \ln (t) + 2 \ln (t + 1) = \ln θ + \ln C$

Where, C is the constant of integration.

$\begin{matrix} \ln {(t + 1)}^{2} - \ln (t^{2}) = \ln + \ln C \\ \ln (\frac{{(t + 1)}^{2}}{t^{2}}) = \ln (θ C) \\ \frac{{(t + 1)}^{2}}{t^{2}} = θ C \cdot \cdot \cdot \cdot \cdot \cdot (6) \end{matrix}$

Put $t = \frac{y}{θ}$ in equation (6),

${(\frac{y + θ}{y})}^{2} = θ C$

Hence, the solution of the given equation $(y^{3} - θ y^{2}) d θ + 2 θ^{2} y d y = 0$ is ${(\frac{y + θ}{y})}^{2} = θ C$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems 33–40, solve the equation given in:
Problem 8.

Step by Step Solution

Step 1: Concept and definition

Step 2: Analyzing the given statement

Step 3: Rewriting the equation (1) in the form dydx=G(yx)

Step 4: Substituting t=yθ in equation (2) to transform the equation in the variables t and θ

Step 5: Find the solution of equation (3) dtdθ=-t(t+1)2θ.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In Problems 33–40, solve the equation given in:Problem 8.

Step by Step Solution

Step 1: Concept and definition

Step 2: Analyzing the given statement

Step 3: Rewriting the equation (1) in the form dydx=G(yx)

Step 4: Substituting t=yθ in equation (2) to transform the equation in the variables t and θ

Step 5: Find the solution of equation (3) dtdθ=-t(t+1)2θ.

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Question: In Problems 33–40, solve the equation given in:
Problem 8.