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 4.

The solution of the given equation is $\frac{- 2}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} (t - 1)}{(x + 3)}) + \ln |(3 {(t - 1)}^{2} + (x + 3)^{2}| = C$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept and definition

Homogeneous differential equation is the equation of the form $f (x, y) dy = g (x, y) dx$ , where the degree of $f (x, y)$ and $g (x, y)$ is same.

Step 2: Analyzing the given statement

One has to solve the equation given in problem 4, i.e.

$(t + x + 2) d x + (3 t - x - 6) d t = 0...... (1)$

Step 3: Transform the equation (1) in homogeneous equation in u and v

Comparing (1) with the equation,

$(a_{1} x + b_{1} y + c_{1}) dx + (a_{2} x + b_{2} y + c_{2}) dy = 0 As a_{1} b_{2} \neq a_{2} b_{1}$

Here, take x as the dependent variable and t as the independent variable.

Let $t = u + h a n d x = v + k$ , where h and k satisfy,

$\begin{array}{l} h + k + 2 = 0...... (2) \\ 3 h - k - 6 = 0...... (3) \end{array}$

Solving (2) and (3),

role="math" localid="1663929399330" $h = 1 and k = - 3$

$Now as t = u + h and x = v + k$

$Therefore, d t = d u and d x = d v$

Substituting these values of x, t, dx, dt in equation (1),

$\begin{matrix} (u + 1 + v - 3 + 2) d x + (3 (u + 1) - (v - 3) - 6) d t = 0 \\ (u + v) d v + (3 u - v) d u = 0 \\ \frac{d v}{d u} = \frac{- (u + v)}{3 u - v} \\ \frac{d v}{d u} = \frac{(\frac{v}{u} + 1)}{1 - 3 (\frac{v}{u})} ...... (4) \end{matrix}$

which is the required homogeneous equation in u and v.

Step 4: Find solution of equation (4)

Let $\frac{v}{u} = z$ in equation (4)

Differentiating both sides,

$\begin{matrix} d z = \frac{u d u - v d v}{u^{2}} \\ v d z = d u - \frac{v}{u} d v \\ v d z = d u - z d v \\ \frac{d v}{d u} = u \frac{d z}{d v} + z \end{matrix}$

Substitute this value of localid="1663929574069" $\frac{dv}{du}$ in equation (4),

$\begin{matrix} u \frac{d z}{d v} + z = \frac{(z + 1)}{1 - 3 z} \\ u \frac{d z}{d v} = \frac{(z + 1)}{1 - 3 z} - z \\ u \frac{d z}{d v} = \frac{z + 1 + 3 z^{2} - z}{1 - 3 z} \\ u \frac{d z}{d v} = \frac{3 z^{2} + 1}{1 - 3 z} \end{matrix}$

Separating the variables,

$\frac{3 z - 1}{- 3 z^{2} - 1} d z = \frac{1}{v} d v$

Integrating both sides,

$\begin{matrix} \int \frac{1 - 3 z}{3 z^{2} + 1} d z = \int \frac{1}{u} d u \\ \int (\frac{1}{3 z^{2} + 1} - \frac{3 z}{3 z^{2} + 1}) d z = \int \frac{1}{u} d u \end{matrix}$

Now, solve the integration part then

$\begin{matrix} \int \frac{1}{3 z^{2} + 1} d z = \frac{1}{\sqrt{3}} \tan^{- 1} \sqrt{3} z \\ 3 \int \frac{z}{3 z^{2} + 1} d z = - \frac{1}{2} \ln |3 z^{2} + 1| \end{matrix}$

Now, equation (4) becomes,

$\begin{matrix} \frac{1}{\sqrt{3}} \tan^{- 1} \sqrt{3} z - \frac{1}{2} \ln |3 z^{2} + 1| = \ln u + C \\ \frac{- 2}{\sqrt{3}} \tan^{- 1} \sqrt{3} z + \ln |3 z^{2} + 1| = - 2 \ln u + C \\ \frac{- 2}{\sqrt{3}} \tan^{- 1} \sqrt{3} z + \ln |3 z^{2} + 1| = - \ln |u^{2}| + C \\ \frac{- 2}{\sqrt{3}} \tan^{- 1} \sqrt{3} z + \ln |3 z^{2} + 1| + \ln |u^{2}| = C \end{matrix}$

$\begin{matrix} \frac{- 2}{\sqrt{3}} \tan^{- 1} \sqrt{3} z + \ln |(3 z^{2} + 1) u^{2}| = C \\ \frac{- 2}{\sqrt{3}} \tan^{- 1} \sqrt{3} (\frac{v}{u}) + \ln |(3 {(\frac{v}{u})}^{2} + 1) u^{2}| = C \\ \frac{- 2}{\sqrt{3}} \tan^{- 1} \sqrt{3} (\frac{t - 1}{x + 3}) + \ln |(3 {(\frac{t - 1}{x + 3})}^{2} + 1) {(x + 3)}^{2}| = C \\ \frac{- 2}{\sqrt{3}} \tan^{- 1} \sqrt{3} (\frac{t - 1}{x + 3}) + \ln |(3 \frac{{(t - 1)}^{2} + {(x + 3)}^{2}}{{(x + 3)}^{2}}) {(x + 3)}^{2}| = C \\ - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} (t - 1}{(x + 3)}) + \ln |3 {(t - 1)}^{2} + {(x + 3)}^{2}| = C \end{matrix}$

Hence, the solution of given equation is $\frac{- 2}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} (t - 1)}{(x + 3)}) + \ln |(3 {(t - 1)}^{2} + (x + 3)^{2}| = C$ .

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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 4.

Step by Step Solution

Step 1: Concept and definition

Step 2: Analyzing the given statement

Step 3: Transform the equation (1) in homogeneous equation in u and v

Step 4: Find solution of equation (4)

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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 4.

Step by Step Solution

Step 1: Concept and definition

Step 2: Analyzing the given statement

Step 3: Transform the equation (1) in homogeneous equation in u and v

Step 4: Find solution of equation (4)

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