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

Step-by-step Solution

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Final answer to the problem

$-\frac{1}{2}\ln\left|y+1\right|+\frac{1}{2}\ln\left|y-1\right|=x+\frac{-1}{e^x}+C_0$
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Step-by-step Solution

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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

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

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

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Learn how to solve problems step by step online. dy/dx=(1+e^(-x))(y^2-1). 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. Apply the formula: b\cdot dy=a\cdot dx\to \int bdy=\int adx, where a=1+e^{-x}, b=\frac{1}{y^2-1}, dyb=dxa=\frac{1}{y^2-1}dy=\left(1+e^{-x}\right)dx, dyb=\frac{1}{y^2-1}dy and dxa=\left(1+e^{-x}\right)dx. Expand the integral \int\left(1+e^{-x}\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. Solve the integral \int\frac{1}{y^2-1}dy and replace the result in the differential equation.

Final answer to the problem

$-\frac{1}{2}\ln\left|y+1\right|+\frac{1}{2}\ln\left|y-1\right|=x+\frac{-1}{e^x}+C_0$

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Function Plot

Plotting: $\frac{dy}{dx}-\left(1+e^{-x}\right)\left(y^2-1\right)$

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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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