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 problem 7-16, solve the equation. $\frac{dy}{dx} = \frac{x}{y^{2} \sqrt{1 + x}}$

The solution of the given differential equation is $y^{3} = 2 [3 x \sqrt{1 + x} - 2 {(1 + x)}^{\frac{3}{2}}] + C$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept of Separable Differential Equation

A first-order ordinary differential equation $\frac{dy}{dx} = f (x, y)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions $g (x)$ that is a function of $x$ alone and $h (y)$ that is a function of $y$ alone.

A separable differential equation can be expressed as $\frac{dy}{dx} = g (x) h (y)$ . By separating the variables, the equation follows $\frac{dy}{h (y)} = g (x) dx$ . Then, on direct integration of both sides, the solution of the differential equation is determined.

Step 2: Solution of the Equation

The given equation is

$\frac{dy}{dx} = \frac{x}{y^{2} \sqrt{1 + x}} \cdot \cdot \cdot \cdot \cdot \cdot (1)$

After separating the variables, equation (1) can be written as

$y^{2} dy = \frac{x}{\sqrt{1 + x}} dx \cdot \cdot \cdot \cdot \cdot \cdot (2)$

Integrate both sides of equation (2). It results,

$\begin{matrix} \int y^{2} dy = \int \frac{x}{\sqrt{1 + x}} dx \\ \int y^{2} dy = 2 \int \frac{x}{2 \sqrt{1 + x}} dx \\ \int y^{2} dy = 2 \int x d (\sqrt{1 + x}) \\ \int y^{2} dy = 2 [x \sqrt{1 + x} - \int \sqrt{1 + x} dx] [\int u dv = uv - \int v du] \end{matrix}$

$\begin{matrix} \frac{y^{3}}{3} = 2 [x \sqrt{1 + x} - \frac{2}{3} {(1 + x)}^{\frac{3}{2}}] + k [k = IntegratingConstant] \\ y^{3} = 2 [3 x \sqrt{1 + x} - 2 {(1 + x)}^{\frac{3}{2}}] + 3 k \\ y^{3} = 2 [3 x \sqrt{1 + x} - 2 {(1 + x)}^{\frac{3}{2}}] + C [C = 3 k = Constant] \end{matrix}$

Therefore, the solution of the given equation is $y^{3} = 2 [3 x \sqrt{1 + x} - 2 {(1 + x)}^{\frac{3}{2}}] + C$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In problem 7-16, solve the equation. $\frac{dy}{dx} = \frac{x}{y^{2} \sqrt{1 + x}}$

Step by Step Solution

Step 1: Concept of Separable Differential Equation

Step 2: Solution of the Equation

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In problem 7-16, solve the equation. dydx=xy21+x

Step by Step Solution

Step 1: Concept of Separable Differential Equation

Step 2: Solution of the Equation

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In problem 7-16, solve the equation. $\frac{dy}{dx} = \frac{x}{y^{2} \sqrt{1 + x}}$