Question

Solve the given differential equation by separation of variables. (ey + 1)2e−y dx + (ex + 1)5e−x dy = 0

Answer 1

Answer:

$\frac{1}{3}(e^y+1)^{-3}=-\frac{1}{6}(e^x+1)^{-6}+C$

Step-by-step explanation:

We are given that differential equation

$(e^y+1)^2e^{-y}dx+(e^x+1)^5e^{-x}dy=0$

$-(e^y+1)^2e^{-y}dx=(e^x+1)^5e^{-x}dy$

$-\frac{dy}{e^{-y}(e^y+1)^2}=\frac{dx}{e^{-x}(e^x+1)^5}$

Taking integration on both sides

$-\int \frac{e^ydy}{(e^y+1)^2}=\int\frac{e^xdx}{(e^x+1)^5}$

Using identity:$\frac{1}{a^x}=a^{-x}$

Substitute $u=1+e^x$

Differentiate w.r.t x

$du=e^xdx$

Substitute $v=1+e^y$

Differentiate w.r.t y

$dv=e^ydy$

Substitute the values then we get

$-\int \frac{dv}{(v)^2}=\int\frac{du}{u^5}$

$-\int v^{-2}dv=\int u^{-5}du$

Using identity:$\frac{1}{a^x}=a^{-x}$

$\frac{1}{3}v^{-3}=-\frac{1}{6}u^{-6}+C$

By using formula :$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

$\frac{1}{3}(\frac{1}{v^3})=-\frac{1}{6}(\frac{1}{u^6})+C$

Substitute the values then we get

$\frac{1}{3}(\frac{1}{(e^y+1)^3})=-\frac{1}{6}(\frac{1}{(e^x+1)^6}+C$

$\frac{1}{3}(e^y+1)^{-3}=-\frac{1}{6}(e^x+1)^{-6}+C$