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.
$(y - x) dx + (x + y) dy = 0$

$y^{2} + 2 xy - x^{2} = C$

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 x₀the 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.

Formulae to be used:

Exactness: If $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ . Otherwise, the equation is not exact.
$\int x^{a} dx = \frac{x^{a + 1}}{a + 1} + C$

Step 2: Given information and simplification

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

Let us check whether the given equation is exact or not.

Then, $M = y - x, N = x + y$ .

Differentiate the value of M and N.

$\begin{array}{l} \frac{\partial M}{\partial y} = 1 \\ \frac{\partial N}{\partial x} = 1 \end{array}$

So, the given equation is exact.

Step 3: Evaluation method

Now, let us assume $M = \frac{\partial F}{\partial x} = y - x$ .

Integrate on both sides.

$\begin{matrix} F = \int (y - x) dx \\ = xy - \frac{x^{2}}{2} + g (y) \end{matrix}$

Differentiate the F with respect to y.

$\begin{matrix} \frac{\partial F}{\partial y} = x + g' (y) \\ = N \end{matrix}$

Equalise the values of N.

$\begin{matrix} x + g' (y) = x + y \\ g' (y) = y \end{matrix}$

Integrate on both sides.

$\begin{matrix} \int g' (y) = \int ydy \\ g (y) = \frac{y^{2}}{2} + C_{1} \end{matrix}$

Substitute in equation of F.

$\begin{matrix} xy - \frac{x^{2}}{2} + \frac{y^{2}}{2} + C_{1} = 0 \\ 2 xy - x^{2} + y^{2} + C_{2} = 0 \\ y^{2} + 2 xy - x^{2} = C \end{matrix}$

Hence, the solution of the given initial value problem is $y^{2} + 2 xy - x^{2} = C$ .

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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.
$(y - x) dx + (x + y) dy = 0$

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.

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Question: In Problems 1-30, solve the equation.y-xdx+x+ydy=0

Step by Step Solution

Step 1: Definition and concepts to be used

Step 2: Given information and simplification

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