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.-
$\frac{dy}{dx} - 4 y = 32 x^{2}$

The solution of the given equation is $y = - 8 x^{2} - 4 x - 1 + {Ce}^{4 x}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information and simplification

Given that, $\frac{dy}{dx} - 4 y = 32 x^{2} \cdot \cdot \cdot \cdot \cdot \cdot (1)$

Let P( x ) = -4.

Find the value of $μ (x)$ .

$\begin{matrix} μ (x) = e^{\int P (x) dx} \\ = e^{- \int 4 dx} \\ = e^{- 4 x} \end{matrix}$

Multiply e^-4x in equation (1) on both sides.

$\begin{matrix} e^{- 4 x} \frac{dy}{dx} - 4 e^{- 4 x} y = 32 e^{- 4 x} x^{2} \\ \frac{d}{dx} [e^{- 4 x} y] = 32 e^{- 4 x} x^{2} \end{matrix}$

Now integrate the equation on both sides.

$\begin{matrix} \int \frac{d}{dx} [e^{- 4 x} y] dx = 32 \int e^{- 4 x} x^{2} dx \\ e^{- 4 x} y = 32 \int e^{- 4 x} x^{2} dx \cdot \cdot \cdot \cdot \cdot \cdot (2) \end{matrix}$

Step 2: Evaluation method

Find the value of $32 \int e^{- 4 x} x^{2} dx$ separately.

Let us take $u = x^{2}, dv = e^{- 4 x} dx$ .

$du = 2 x dx, v = - \frac{e^{- 4 x}}{4}$ .

Use the integration by parts formula.

$\begin{array}{l} 32 \int e^{- 4 x} x^{2} dx = 32 [\frac{- x^{2} e^{- 4 x}}{4} + \int \frac{{xe}^{- 4 x}}{2} dx] \\ 32 \int e^{- 4 x} x^{2} dx = 32 [\frac{- x^{2} e^{- 4 x}}{4} + \frac{1}{2} \int {xe}^{- 4 x} dx] \cdot \cdot \cdot \cdot \cdot \cdot (3) \end{array}$

Again integrate $\int {xe}^{- 4 x} dx$ separately.

$\begin{array}{l} u = x, dv = e^{- 4 x} \\ du = dx, v = - \frac{e^{- 4 x}}{4} \end{array}$

Use the integration by parts method.

$\begin{matrix} \int {xe}^{- 4 x} dx = - \frac{{xe}^{- 4 x}}{4} + \frac{1}{4} \int e^{- 4 x} dx \\ = - x \frac{e^{- 4 x}}{4} + \frac{1}{4} [- \frac{e^{- 4 x}}{4}] + C_{1} \end{matrix}$

Substitute in equation (3)

$\begin{matrix} 32 \int e^{- 4 x} x^{2} dx = 32 [\frac{- x^{2} e^{- 4 x}}{4} + \frac{1}{2} [- x \frac{e^{- 4 x}}{4} + \frac{1}{4} [- \frac{e^{- 4 x}}{4}] + C_{1}]] \\ = 32 [- x^{2} \frac{e^{- 4 x}}{4} - x \frac{e^{- 4 x}}{8} - \frac{e^{- 4 x}}{32} + C] \\ = - 8 x^{2} e^{- 4 x} - 4 {xe}^{- 4 x} - e^{- 4 x} + C \end{matrix}$

Now substitute in equation (2)

$\begin{matrix} e^{- 4 x} y = 32 \int e^{- 4 x} x^{2} dx \\ e^{- 4 x} y = - 8 x^{2} e^{- 4 x} - 4 {xe}^{- 4 x} - e^{- 4 x} + C \\ y = \frac{- 8 x^{2} e^{- 4 x} - 4 {xe}^{- 4 x} - e^{- 4 x} + C}{e^{- 4 x}} \\ y = - 8 x^{2} - 4 x - 1 + {Ce}^{4 x} \end{matrix}$

So, the solution is data-custom-editor="chemistry" $y = - 8 x^{2} - 4 x - 1 + {Ce}^{4 x}$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Question: In Problems 1-30, solve the equation.-
$\frac{dy}{dx} - 4 y = 32 x^{2}$

Step by Step Solution

Step 1: Given information and simplification

Step 2: Evaluation method

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Question: In Problems 1-30, solve the equation.- dydx-4y=32x2

Step by Step Solution

Step 1: Given information and simplification

Step 2: Evaluation method

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Question: In Problems 1-30, solve the equation.-
$\frac{dy}{dx} - 4 y = 32 x^{2}$