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 33-40 Solve the equation given in Problem 3.
$\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}$

$y = \frac{- 3}{x^{4} + Cx} and y \equiv 0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given equation solve by Bernoulli Equations.

A first-order equation that can be written in the form,

$\frac{dy}{dx} + P (x) y = Q (x) y^{n}$ ,

where P(x) and Q(x) are continuous on an interval (a, b) and n is a real number, is called a Bernoulli equation.

Step 2: Solve the given equation

The given equation is,

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

Both sides divide by y² in the above equation,

$y^{- 2} \frac{dy}{dx} + \frac{y^{- 1}}{x} = x^{3} ...... (1)$

Let,

$\begin{matrix} t = y^{- 1} \\ \frac{dt}{dx} = - y^{- 2} \frac{dy}{dx} \end{matrix}$

Substitute the $t = y^{- 1} and \frac{dt}{dx} = - y^{- 2} \frac{dy}{dx}$ in the equation (1),

$\begin{matrix} - \frac{dt}{dx} + \frac{t}{x} = x^{3} \\ \frac{dt}{dx} - \frac{t}{x} = - x^{3} ...... (2) \end{matrix}$

Step 3: Compare equation (2) with the standard form of the linear equation,

The standard form of the linear equation is,

$\frac{dy}{dx} + P (x) y = Q (x)$

One has,

$P (x) = - \frac{1}{x}$

Now find the integrating factor,

$\begin{matrix} μ (x) = e^{\int P (x) dx} \\ = e^{- \int \frac{1}{x} dx} \\ = e^{- \log (x)} \\ = \frac{1}{x} \end{matrix}$

Both sides Multiply by $μ (x) = \frac{1}{x}$ in the equation (2),

$\frac{1}{x} \frac{dt}{dx} - (\frac{1}{x}) \frac{t}{x} = - (\frac{1}{x}) x^{3}$

Simplify the above equation,

$\begin{matrix} \frac{1}{x} \frac{dt}{dx} - \frac{t}{x^{2}} = - x^{2} \\ \frac{d}{dx} (\frac{t}{x}) = - x^{2} \end{matrix}$

Integrate both sides and solve for the solution.

$\frac{t}{x} = - \int x^{2} dx + c$

Substitute t = y^-1 and simplify the above equation,

$\begin{matrix} \frac{y^{- 1}}{x} = - \frac{x^{3}}{3} + c \\ \frac{1}{y} = - \frac{x^{4}}{3} + cx \\ \frac{1}{y} = \frac{x^{4} + 3 cx}{- 3} \end{matrix}$

Where, C = -3c

$y = \frac{- 3}{x^{4} + Cx}$

Hence, the solve the given equation is $y = \frac{- 3}{x^{4} + Cx} and y \equiv 0$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In problems 33-40 Solve the equation given in Problem 3.
$\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}$

Step by Step Solution

Step 1: Given equation solve by Bernoulli Equations.

Step 2: Solve the given equation

Step 3: Compare equation (2) with the standard form of the linear equation,

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In problems 33-40 Solve the equation given in Problem 3.dydx+yx=x3y2

Step by Step Solution

Step 1: Given equation solve by Bernoulli Equations.

Step 2: Solve the given equation

Step 3: Compare equation (2) with the standard form of the linear equation,

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Question: In problems 33-40 Solve the equation given in Problem 3.
$\frac{dy}{dx} + \frac{y}{x} = x^{3} y^{2}$