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 , solve the equation.
$\frac{dy}{dx} = {(x + y + 1)}^{2} - {(x + y - 1)}^{2}$

The solution of the given equation is $y = \frac{{Ce}^{4 x} - 4 x - 1}{4}$

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.

Formulae to be used:

Integration by parts: $\int udv = uv - \int vdu$ .
$a^{2} - b^{2} = (a - b) (a + b)$
$\int e^{ax} dx = \frac{e^{ax}}{a} + C$ .

Step 2: Given information and simplification

Given that, $\frac{dy}{dx} = {(x + y + 1)}^{2} - {(x + y - 1)}^{2} ...... (1)$

Evaluate the equation (1).

$\begin{matrix} \frac{dy}{dx} = {(x + y + 1)}^{2} - {(x + y - 1)}^{2} \\ \frac{dy}{dx} = (x + y + 1 - (x + y - 1)) (x + y + 1 + (x + y - 1)) \\ = 2 (2 x + 2 y) \\ = 4 (x + y) \end{matrix}$

Let us take $t = x + y \to \frac{dy}{dx} = \frac{dt}{dx} - 1$ .

Then,

$\begin{matrix} \frac{dt}{dx} - 1 = 4 t \\ \frac{dt}{dx} = 4 t + 1 \\ \frac{dt}{4 t + 1} = dx \end{matrix}$

Step 3: Evaluation method

Now integrate the equation (2) on both sides.

$\begin{matrix} \int \frac{dt}{4 t + 1} = \int dx \\ \frac{\ln |4 t + 1|}{4} = x + C_{1} \\ \ln |4 t + 1| = 4 x + C \end{matrix}$

Simplify it.

$\begin{matrix} 4 t + 1 = {Ce}^{4 x} \\ 4 t = {Ce}^{4 x} - 1 \\ t = \frac{{Ce}^{4 x} - 1}{4} \end{matrix}$

Substitute the value of t.

$\begin{matrix} x + y = \frac{{Ce}^{4 x} - 1}{4} \\ y = \frac{{Ce}^{4 x} - 1}{4} - x \\ y = \frac{{Ce}^{4 x} - 4 x - 1}{4} \end{matrix}$

Hence, the solution of the given initial value problem is $y = \frac{{Ce}^{4 x} - 4 x - 1}{4}$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems , solve the equation.
$\frac{dy}{dx} = {(x + y + 1)}^{2} - {(x + y - 1)}^{2}$

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 , solve the equation.
$\frac{dy}{dx} = {(x + y + 1)}^{2} - {(x + y - 1)}^{2}$