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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 2: Q 1RP (page 79)

Question: In Problems 1 - 30, solve the equation.
$鈥� \frac{dy}{dx} = \frac{e^{x + y}}{y - 1}$

Short Answer

Expert verified

$e^{x} + {ye}^{- y} = C$

Step by step solution

01

Given information and simplification

Given that, $\frac{dy}{dx} = \frac{e^{x + y}}{y - 1} 鈰� 鈰� 鈰� 鈰� 鈰� 鈰� (1)$

Evaluate the equation (1).

$\begin{matrix} \frac{dy}{dx} = \frac{e^{x + y}}{y - 1} \\ \frac{dy}{dx} = \frac{e^{x} e^{y}}{y - 1} \\ = \frac{e^{x}}{(y - 1) e^{- y}} \\ (y - 1) e^{- y} dy = e^{x} dx 鈰� 鈰� 鈰� 鈰� 鈰� 鈰� (2) \end{matrix}$

Now integrate the equation (2) on both sides.

$\begin{matrix} 鈭� (y - 1) e^{- y} dy = 鈭� e^{x} dx \\ 鈭� (y - 1) e^{- y} dy = e^{x} + C \end{matrix}$

02

Evaluation method

Find the value of $鈭� (y - 1) e^{- y} dy$ separately.

Let us take $u = y - 1, dv = e^{- y} dy$ .

$du = 1 dy, v = - e^{- y}$ .

Use the integration by parts formula.

$\begin{matrix} 鈭� (y - 1) e^{- y} dy = - e^{- y} (y - 1) + 鈭� e^{- y} dy \\ = - {ye}^{- y} + e^{- y} - e^{- y} \\ = - {ye}^{- y} \end{matrix}$

Then,

$\begin{matrix} 鈭� (y - 1) e^{- y} dy = e^{x} + C \\ - {ye}^{- y} = e^{x} + C \\ e^{x} + 鈥� {ye}^{- y} = C \end{matrix}$

So, the solution is $e^{x} + {ye}^{- y} = C$

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Most popular questions from this chapter

Question: In Problems 1-30, solve the equation.

$鈥夆赌� \frac{dy}{d 胃} + 2 y = y^{2}$

Question: In Problems 1-30, solve the equation.

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

Use the method discussed under 鈥淓quations of the Form $\frac{dy}{dx} = G (ax + by)$ 鈥� to solve problems 17-20.

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

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

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

$胃诲测 - 测诲胃 = \sqrt{胃测诲胃}$

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