3) please show work 3. ВCSolve e У(у'+1)%-хе". Hint: Letz=x+y

Given: e-yy'+1=xex So, we can also write as: e-yy'+1=xexdydx+1=xexe-ydydx+1=xex--y...We know…

Answered: 3. ВCSolve e(у'+1)-хе". Hint: Let z=x+y

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

3) please show work

3. ВCSolve e У(у'+1)%-хе". Hint: Let
z=x+y

Expert Solution

Step 1

Given:

$e^{- y} (y' + 1) = x e^{x}$

So, we can also write as:

$e^{- y} (y' + 1) = x e^{x} (\frac{d y}{d x} + 1) = x \frac{e^{x}}{e^{- y}} (\frac{d y}{d x} + 1) = x e^{x - (- y)} . . . (W e k n o w \frac{a^{m}}{a^{n}} = a^{m - n}) (\frac{d y}{d x} + 1) = x e^{x + y} . . . . . (1)$

Let,

$\begin{array}{rcl} z & = & x + y \\ S o, \\ \frac{d}{d x} (z) & = & \frac{d}{d x} (x) + \frac{d}{d x} (y) \\ \frac{d z}{d x} & = & 1 + \frac{d y}{d x} \\ (\frac{d z}{d x} - 1) & = & \frac{d y}{d x} \end{array}$