Solve the separable equation −ex+dy(x)dx(ex+1)y(x)=0: Solve for dy(x)dx: dy(x)dx=ex(ex+1)y(x) INTERMEDIATE STEPS: Solve for dy(x)dx: y(x)dy(x)dx(ex+1)−ex=0 Hint: Isolate terms with dy(x)dx to the left hand side. Add ex to both sides: y(x)dy(x)dx(ex+1)=ex Solve for dy(x)dx. Divide both sides by (ex+1)y(x): dy(x)dx=exy(x)(ex+1) Multiply both sides by y(x): dy(x)dxy(x)=exex+1 Integrate both sides with respect to x: ∫dy(x)dxy(x)dx=∫exex+1dx Evaluate the integrals: y(x)22=log⁡(ex+1)+c1, where c1 is an arbitrary constant. INTERMEDIATE STEPS: Take the integral: integral exex+1dx For the integrand exex+1, substitute u=ex+1anddu=exdx: =∫1udu The integral of

Find out the answer to "(1+ex)ydy−exdx=0"

beljuA

Answered question

2021-09-06

$(1 + e^{x}) y d y - e^{x} d x = 0$

Answer & Explanation

yunitsiL

Skilled2021-09-07Added 108 answers

Solve the separable equation $- e^{x} + \frac{d y (x)}{d x} (e^{x} + 1) y (x) = 0 :$
Solve for $\frac{d y (x)}{d x} :$
$\frac{d y (x)}{d x} = \frac{e^{x}}{(e^{x} + 1) y (x)}$
INTERMEDIATE STEPS:
Solve for $\frac{d y (x)}{d x} :$
$y (x) \frac{d y (x)}{d x} (e^{x} + 1) - e^{x} = 0$
Hint: Isolate terms with $\frac{d y (x)}{d x}$ to the left hand side.
Add $e^{x}$ to both sides:
$y (x) \frac{d y (x)}{d x} (e^{x} + 1) = e^{x}$
Solve for $\frac{d y (x)}{d x} .$
Divide both sides by $(e^{x} + 1) y (x) :$
$\frac{d y (x)}{d x} = \frac{e^{x}}{y (x) (e^{x} + 1)}$
Multiply both sides by y(x):
$\frac{d y (x)}{d x} y (x) = \frac{e^{x}}{e^{x} + 1}$
Integrate both sides with respect to x:
$\int \frac{d y (x)}{d x} y (x) d x = \int \frac{e^{x}}{e^{x} + 1} d x$
Evaluate the integrals:
$y \frac{{(x)}^{2}}{2} = \log (e^{x} + 1) + c_{1}$ , where $c_{1}$ is an arbitrary constant.
INTERMEDIATE STEPS:
Take the integral:
integral $\frac{e^{x}}{e^{x} + 1} d x$
For the integrand $\frac{e^{x}}{e^{x} + 1}$ , substitute $u = e^{x} + 1 and d u = e^{x} d x :$
$= \int \frac{1}{u} d u$
The integral of

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