# Solve the given differential equation. If an initial condition is

Solve the given differential equation. If an initial condition is given, also find the solution that satisfies it.
$\left({e}^{x}+1\right)\frac{dy}{dx}=y-y{e}^{x}$
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Edward Patten
Step 1
Given differential equation,
$\left({e}^{x}+1\right)\frac{dy}{dx}=y-y{e}^{x}$
Step 2
Since we have,
$\left({e}^{x}+1\right)\frac{dy}{dx}=y-y{e}^{x}$
$\left({e}^{x}+1\right)\frac{dy}{dx}=y\left(1-{e}^{x}\right)$
$\frac{dy}{y}=\frac{1-{e}^{x}}{1+{e}^{x}}dx$ [Variable seperable form]
Integrate both sides,
$\int \frac{dy}{y}=\int \frac{1-{e}^{x}}{1+{e}^{x}}dx$ (1)
Step 3
Now firstly simplify $\int \frac{1-{e}^{x}}{1+{e}^{x}}dx$,
Substitute ${e}^{x}=u⇒{e}^{x}dx=du⇒dx=\frac{1}{{e}^{x}}du$, so
$\int \frac{1-{e}^{x}}{1+{e}^{x}}dx=\int \frac{1-u}{1+u}\frac{du}{u}$
$=\int \frac{1-u}{u\left(1-u\right)}du$ (2)
Simplify $\frac{1-u}{u\left(1+u\right)}$ using partial fraction method,
$\frac{1-u}{u\left(1+u\right)}=\frac{A}{u}+\frac{B}{1+u}$ (3)
$1-u=A\left(1+u\right)+Bu$
put $u=0,A=1$
Put $u=-1,B=-2$
Substitute the value of A and B in (3),
$\frac{1-u}{u\left(1+u\right)}=\frac{1}{u}-\frac{2}{1+u}$ (4)
Using (4) i (2),
$\int \frac{1-{e}^{x}}{1+{e}^{x}}=\int \left(\frac{1}{u}-\frac{2}{1+u}\right)du$
###### Did you like this example?
Mollie Nash
Given differential equation is $\left({e}^{x}+1\right)\frac{dy}{dx}=y-y{e}^{x}$
$\left({e}^{x}+1\right)\frac{dy}{dx}=y\left(1-{e}^{x}\right)$
on separaing the variables, we get
$\frac{dy}{y}=\frac{1-{e}^{x}}{1+{e}^{x}}dx$
Integrating on both sides
$\int \frac{dy}{y}=\int \frac{1-{e}^{x}}{1+{e}^{x}}dx$
$\mathrm{ln}y=\int \frac{\left(1+{e}^{x}\right)-2{e}^{x}}{1+{e}^{x}}dx$
$\mathrm{ln}y=\int \left(1-2\frac{{e}^{x}}{1+{e}^{x}}\right)dx$
$\mathrm{ln}y=x-2\mathrm{ln}\left(1+{e}^{x}\right)+c$
$y={e}^{x-2h\left(1+e\right)+c}$
$y={e}^{x}{e}^{-2h\left(1+e\right)}{e}^{c}$

$y={c}_{1}{e}^{x}{\left(1+{e}^{x}\right)}^{-2}$
Therefore the solution of the given differential equation is $y={c}_{1}\frac{{e}^{x}}{{\left(1+{e}^{x}\right)}^{2}}$
karton