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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 79

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.
$y (x - y - 2) dx + x (y - x + 4) dy = 0$

$x^{2} y^{- 2} - 2 {xy}^{- 1} - 4 {xy}^{- 2} = C and y \equiv 0$

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.

Exactness: If $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ . Otherwise, the equation is not exact.

Special integrating factor: $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N}, μ (x) = e^{\int P (x) dx}$ .

Formulae to be used:

$\int x^{a} dx = \frac{x^{a + 1}}{a + 1} + C$ .
$\int e^{ax} dx = \frac{e^{ax}}{a} + C$ .

Step 2: Given information and simplification

Given that, $y (x - y - 2) dx + x (y - x + 4) dy = 0 ...... (1)$

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

Then, $M = y (x - y - 2), N = x (y - x + 4)$ .

Differentiate the value of M and N.

$\begin{array}{l} \frac{\partial M}{\partial y} = x - 2 y - 2 \\ \frac{\partial N}{\partial x} = y - 2 x + 4 \end{array}$

So, the given equation is not exact. Then, find the special integrating factor,

$\begin{matrix} \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{y - 2 x + 4 - x + 2 y + 2}{y (x - y - 2)} \\ = \frac{3 y - 3 x + 6}{y (x - y - 2)} \\ = \frac{- 3 (x - y - 2)}{y (x - y - 2)} \\ = - \frac{3}{y} \end{matrix}$

Step 3: Find the exactness

Find the value of $μ (y)$ .

$\begin{matrix} μ (y) = e^{\int - \frac{3}{y} dy} \\ = e^{- 3 \ln |y|} \\ = y^{- 3} \end{matrix}$

Multiply y^--3 in equation (1) on both sides.

$y^{- 2} (x - y - 2) dx + {xy}^{- 3} (y - x + 4) dy = 0$

Now again check whether the founded equation is exact or not.

$M = y^{- 2} (x - y - 2), N = {xy}^{- 3} (y - x + 4)$

Differentiate the value of M and N.

$\begin{array}{l} \frac{\partial M}{\partial y} = \frac{4}{y^{3}} + \frac{1}{y^{2}} - \frac{2 x}{y^{3}} \\ \frac{\partial N}{\partial x} = \frac{4}{y^{3}} + \frac{1}{y^{2}} - \frac{2 x}{y^{3}} \end{array}$

Therefore, the founded equation is exact.

Step 4: Evaluation method

Now, let us assume $M = \frac{\partial F}{\partial x} = y^{- 2} (x - y - 2)$ .

Integrate on both sides.

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

Differentiate the F with respect to y.

$\begin{matrix} \frac{\partial F}{\partial y} = \frac{4 x}{y^{3}} - \frac{x^{2}}{y^{3}} + \frac{x}{y^{2}} + g' (y) \\ = N \end{matrix}$

Equalize the values of N.

$\begin{matrix} \frac{4 x}{y^{3}} - \frac{x^{2}}{y^{3}} + \frac{x}{y^{2}} + g' (y) = \frac{4 x}{y^{3}} - \frac{x^{2}}{y^{3}} + \frac{x}{y^{2}} \\ g' (y) = 0 \end{matrix}$

Integrate on both sides.

$\begin{matrix} \int g' (y) = \int 0 dy \\ g (y) = C_{1} \end{matrix}$

Substitute in the equation of F.

$\frac{x^{2}}{2 y^{2}} - \frac{x}{y} - \frac{2 x}{y^{2}} + C_{1} = 0$

Multiply 2 on both sides.

$\begin{matrix} x^{2} y^{- 2} - 2 {xy}^{- 1} - 4 {xy}^{2} + C_{2} = 0 \\ x^{2} y^{- 2} - 2 {xy}^{- 1} - 4 {xy}^{2} = C \end{matrix}$

Hence, the solution of the given initial value problem is $x^{2} y^{- 2} - 2 {xy}^{- 1} - 4 {xy}^{2} = C and y \equiv 0$

Most popular questions for Math Textbooks

In problem $1 - 6$ , determine whether the given differential equation is separable $s^{2} + \frac{d s}{d t} = \frac{s + 1}{s t}$ .

Question: Coupled Equations. In analyzing coupled equations of the form

$\begin{array}{l} \frac{dy}{dt} = ax + by \\ \frac{dx}{dt} = αx + βy \end{array}$

where a, b, $α and y$ are constants, we may wish to determine the relationship between x and y rather than the individual solutions x(t), y(t). For this purpose, divide the first equation by the second to obtain

$\frac{dy}{dx} = \frac{ax + by}{αx + βy}$

This new equation is homogeneous, so we can solve it via the substitution $v = \frac{x}{y}$ . We refer to the solutions of (17) as integral curves. Determine the integral curves for the system

$\begin{array}{l} \frac{dy}{dt} = - 4 x - y \\ \frac{dx}{dt} = 2 x - y \end{array}$

In problem $1 - 6$ , determine whether the differential equation is separable role="math" localid="1654775979001" $(x y^{2} + 3 y^{2}) d y - 2 x d x = 0$ .

Question: Riccati Equation. An equation of the form (18) $\frac{d y}{d x} = P (x) y^{2} + Q (x) y + R (x)$ is called a generalized Riccati equation.

If one solution—say, u(x)—of (18) is known, show that the substitution $y = u + \frac{1}{v}$ reduces (18) to a linear equation in v.
Given that $u (x) = x$ is a solution to $\frac{d y}{d x} = x^{3} {(y - x)}^{2} + \frac{y}{x},$

use the result of part (a) to find all the other solutions to this equation. (The particular solution $u (x) = x$ can be found by inspection or by using a Taylor series method; see Section 8.1.)

Question: In problems 33-40 Solve the equation given in Problem 3.

$\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}$

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