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Q2.6-1E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 76

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

In problems 1-8 identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form $y^{'} = G (ax + by)$ . $2 t \times dx + (t^{2} - x^{2}) dt = 0$

The given equation is the form of both homogeneous and Bernoulli.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

General form of homogeneous, Bernoulli, linear coefficients of the form of y'=Gax+by

Homogeneous equation

If the right-hand side of the equation $\frac{d y}{d x} = f (x, y)$ can be expressed as a function of the ratio $\frac{y}{x}$ alone, then we say the equation is homogeneous.

Equations of the form $\frac{d y}{d x} = G (a x + b y)$ When the right-hand side of the equation $\frac{d y}{d x} = f (x, y)$ can be expressed as a function of the combination $(a x + b y)$ where and are constants, that is, $\frac{d y}{d x} = G (a x + b y)$ then the substitution $Z = a x + b y$ transforms the equation into a separable one.

Bernoulli’s equation

A first-order equation that can be written in the form $\frac{d y}{d x} + P (x) y = Q (x) y^{n}$ , where $P (x)$ and $Q (x)$ are continuous on interval $(a, b)$ and localid="1654934017689" $n$ is a real number, is called a Bernoulli equation.

Equation of Linear coefficients

We have used various substitutions for $y$ to transform the original equation into a new equation that we could solve. In some cases, we must transform both $x$ and $y$ into new variables, say $u$ and $v$ . This is the situation for equations with linear coefficients-that is, equations of the form $(a_{1} x + b_{1} y + c_{1}) d x + (a_{2} x + b_{2} y + c_{2}) d y = 0$

Evaluate the given equation

Given, $2 t x d x + (t^{2} - x^{2}) d t = 0$

By Evaluation,

role="math" localid="1654934940793" $\begin{array}{rcl} 2 t x d x + (t^{2} - x^{2}) d t & = & 0 \\ 2 t x d x & = & - (t^{2} - x^{2}) d t \\ \frac{d x}{d t} & = & \frac{t^{2} - x^{2}}{2 t x} \\ = & - \frac{1}{2} (\frac{t}{x} - \frac{x}{t}) \\ = & \frac{1}{2} (\frac{x}{t} - \frac{1}{\frac{x}{t}}) \end{array}$ .........(1)

Substitution method

Let us consider $v = \frac{x}{t}$

From Equation (1),

role="math" localid="1654934950153" $\begin{array}{rcl} \frac{x^{2} - t^{2}}{2 t x} & = & G (\frac{x}{t}) \\ = & \frac{1}{2} (\frac{x}{t} - \frac{1}{\frac{x}{t}}) \\ G (v) & = & \frac{1}{2} (v - \frac{1}{v}) \end{array}$

So, the given equation is homogeneous.

It can be rewritten as,

$\begin{array}{rcl} \frac{d x}{d t} & = & \frac{x^{2} - t^{2}}{2 t x} \\ = & \frac{x}{2 t} - \frac{t}{2 x} \\ \frac{d x}{d t} + (- \frac{1}{2 t}) x & = & (- \frac{t}{2}) x^{- 1} \end{array}$

Comparing the found equation with Bernoulli general form of equation. It seems that the given equation is Bernoulli equation as well.

Therefore, the given equation is the form of both homogeneous and Bernoulli.

Most popular questions for Math Textbooks

Use the method discussed under “Homogeneous Equations” to solve problems 9-16 $(xy + y^{2}) dx - x^{2} dy = 0$

Question: In Problems , solve the equation.

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

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.

$(3 x^{2} - y^{2}) dx + (xy - x^{3} y^{- 1}) dy = 0$

Question: In Problems 33–40, solve the equation given in: Problem 7.

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

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