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 1-30, solve the equation.
$(3 x - y - 5) dx + (x - y + 1) dy = 0$

$[{(y - 4)}^{2} - 3 {(x - 3)}^{2}] {|\frac{\sqrt{3} (x - 3) + (y - 4)}{\sqrt{3} (x - 3) - (y - 4)}|}^{\frac{1}{\sqrt{3}}} = C$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition and concepts to be used

Definition of Initial Value Problem: By an initial value problem for an nth-order differential equation $F (x, y, \frac{dy}{dx}, ..., \frac{d^{n} y}{{dx}^{n}}) = 0$ we mean: Find a solution to the differential equation on an interval I that satisfies at x₀ the n initial conditions

$\begin{matrix} y (x_{0}) = y_{0}, \\ \frac{dy}{dx} (x_{0}) = y_{1}, \\ . \\ . \\ . \\ \frac{d^{n - 1} y}{{dx}^{n - 1}} (x_{0}) = y_{n - 1},, \end{matrix}$

Where $x_{0} \in I$ and $y_{0}, y_{1}, ..., y_{n - 1}$ are given constants.

Homogeneous: it is satisfying only the function turns into $\frac{y}{x}$ format.

Linear coefficient: here we need to convert both the variables into another two variables. For example, x as u and y as v.

Formulae to be used:

Integration by parts: $\int udv = uv - \int vdu$ .
$\int x^{a} dx = \frac{x^{a + 1}}{a + 1} + C$ .
$\int \frac{1}{x} dx = \ln |x| + C$ .

Step 2: Given information and simplification

Given that, $(3 x - y - 5) dx + (x - y + 1) dy = 0 ...... (1)$

Since the given equation is linear coefficients. So, we can take

$\begin{array}{l} x = u + h \to dx = du \\ y = v + k \to dy = dv \end{array}$

Evaluate the equation (1).

$\begin{matrix} (3 x - y - 5) dx + (x - y + 1) dy = 0 \\ \frac{dy}{dx} = \frac{3 x - y - 5}{x - y + 1} \\ \frac{dv}{du} = \frac{3 (u + h) - (v + k) - 5}{u + h - v - k + 1} \\ = \frac{3 u - v + 3 h - k - 5}{u - v + h - k + 1} \end{matrix}$

Using the condition, we get,

$\begin{matrix} 3 h - k - 5 = 0 \\ h - k + 1 = 0 \end{matrix}$

Solve the above equations to find the values of h and k.

$\begin{matrix} 2 h - 6 = 0 \\ 2 h = 6 \\ h = 3 \end{matrix}$

Substitute h = 3.

$\begin{matrix} 3 - k + 1 = 0 \\ k = 4 \end{matrix}$

Then,

$\begin{matrix} \frac{dv}{du} = \frac{3 u - v}{u - v} \\ = \frac{u (3 - \frac{v}{u})}{u (1 - \frac{v}{u})} \\ = \frac{3 - \frac{v}{u}}{1 - \frac{v}{u}} \end{matrix}$

So, the founded equation is homogeneous.

Let us take $s = \frac{v}{u} \to \frac{dv}{du} = s + u \frac{ds}{du}$ .

Then,

$\begin{matrix} s + u \frac{ds}{du} = \frac{3 - s}{1 - s} \\ u \frac{ds}{du} = \frac{3 - s}{1 - s} - s \\ = \frac{3 - s - s + s^{2}}{1 - s} \\ = \frac{3 - 2 s + s^{2}}{1 - s} \\ \frac{1 - s}{s^{2} - 2 s + 3} ds = \frac{du}{u} \end{matrix}$

Step 3: Evaluation method

Now integrate the equation (2) on both sides.

$\int \frac{1 - s}{s^{2} - 2 s + 3} ds = \int \frac{du}{u}$

Find the value of $\int \frac{1 - s}{s^{2} - 2 s + 3} ds$ separately.

Let $t = s^{2} - 2 s + 3 \to dt = - 2 (1 - s) ds$ .

Then,

$\begin{matrix} \int \frac{1 - s}{s^{2} - 2 s + 3} ds = - \int \frac{1}{t} \frac{dt}{2} \\ = - \frac{1}{2} \int \frac{1}{t} dt \\ = - \frac{1}{2} \ln |t| \\ = - \frac{1}{2} \ln |s^{2} - 2 s + 3| \end{matrix}$

Substitute the value here.

$\begin{matrix} \int \frac{1 - s}{s^{2} - 2 s + 3} ds = \int \frac{du}{u} \\ - \frac{1}{2} \ln |s^{2} - 2 s + 3| = \ln |u| + C_{1} \\ \ln |s^{2} - 2 s + 3| = - 2 \ln |u| + C_{2} \\ = - \ln |u^{2}| + C_{2} \\ (s^{2} - 2 s + 3) u^{2} = C_{3} \end{matrix}$

Again, substitute the value of s, u and v

$\begin{matrix} ({(\frac{v}{u})}^{2} - 2 (\frac{v}{u}) + 3) u^{2} = C_{2} \\ (\frac{v^{2}}{u^{2}} - 2 \frac{v}{u} + 3) u^{2} = C_{2} \\ \frac{v^{2} - 2 uv + 3 u^{2}}{u^{2}} u^{2} = C_{2} \\ v^{2} - 2 uv + 3 u^{2} = C_{2} \end{matrix}$

$[{(y - 4)}^{2} - 3 {(x - 3)}^{2}] {|\frac{\sqrt{3} (x - 3) + (y - 4)}{\sqrt{3} (x - 3) - (y - 4)}|}^{\frac{1}{\sqrt{3}}} = C$

Hence, the solution of given initial value problem is .

$[{(y - 4)}^{2} - 3 {(x - 3)}^{2}] {|\frac{\sqrt{3} (x - 3) + (y - 4)}{\sqrt{3} (x - 3) - (y - 4)}|}^{\frac{1}{\sqrt{3}}} = C$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems 1-30, solve the equation.
$(3 x - y - 5) dx + (x - y + 1) dy = 0$

Step by Step Solution

Step 1: Definition and concepts to be used

Step 2: Given information and simplification

Step 3: Evaluation method

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In Problems 1-30, solve the equation.3x-y-5dx+x-y+1dy=0

Step by Step Solution

Step 1: Definition and concepts to be used

Step 2: Given information and simplification

Step 3: Evaluation method

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Question: In Problems 1-30, solve the equation.
$(3 x - y - 5) dx + (x - y + 1) dy = 0$