$e^y\left(\frac{dy}{dx}+1\right)=1$

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Final answer to the problem

$\ln\left|1-e^y\right|=-x+C_0$
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Step-by-step Solution

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Apply the formula: $x\left(a+b\right)$$=xa+xb$, where $a=\frac{dy}{dx}$, $b=1$, $x=e^y$ and $a+b=\frac{dy}{dx}+1$

$e^y\frac{dy}{dx}+e^y=1$

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$e^y\frac{dy}{dx}+e^y=1$

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Learn how to solve problems step by step online. e^y(dy/dx+1)=1. Apply the formula: x\left(a+b\right)=xa+xb, where a=\frac{dy}{dx}, b=1, x=e^y and a+b=\frac{dy}{dx}+1. Apply the formula: a\frac{b}{c}=\frac{ba}{c}, where a=e^y, b=dy and c=dx. Apply the formula: x+a=b\to x=b-a, where a=e^y, b=1, x+a=b=\frac{e^ydy}{dx}+e^y=1, x=\frac{e^ydy}{dx} and x+a=\frac{e^ydy}{dx}+e^y. Group the terms of the differential equation. Move the terms of the y variable to the left side, and the terms of the x variable to the right side of the equality.

Final answer to the problem

$\ln\left|1-e^y\right|=-x+C_0$

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Function Plot

Plotting: $e^y\left(\frac{dy}{dx}+1\right)-1$

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1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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