Answer:
[tex]\frac{1}{3}(e^y+1)^{-3}=-\frac{1}{6}(e^x+1)^{-6}+C[/tex]
Step-by-step explanation:
We are given that differential equation
[tex](e^y+1)^2e^{-y}dx+(e^x+1)^5e^{-x}dy=0[/tex]
[tex]-(e^y+1)^2e^{-y}dx=(e^x+1)^5e^{-x}dy[/tex]
[tex]-\frac{dy}{e^{-y}(e^y+1)^2}=\frac{dx}{e^{-x}(e^x+1)^5}[/tex]
Taking integration on both sides
[tex]-\int \frac{e^ydy}{(e^y+1)^2}=\int\frac{e^xdx}{(e^x+1)^5}[/tex]
Using identity:[tex]\frac{1}{a^x}=a^{-x}[/tex]
Substitute [tex]u=1+e^x[/tex]
Differentiate w.r.t x
[tex]du=e^xdx[/tex]
Substitute [tex]v=1+e^y[/tex]
Differentiate w.r.t y
[tex]dv=e^ydy[/tex]
Substitute the values then we get
[tex]-\int \frac{dv}{(v)^2}=\int\frac{du}{u^5}[/tex]
[tex]-\int v^{-2}dv=\int u^{-5}du[/tex]
Using identity:[tex]\frac{1}{a^x}=a^{-x}[/tex]
[tex]\frac{1}{3}v^{-3}=-\frac{1}{6}u^{-6}+C[/tex]
By using formula :[tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]
[tex]\frac{1}{3}(\frac{1}{v^3})=-\frac{1}{6}(\frac{1}{u^6})+C[/tex]
Substitute the values then we get
[tex]\frac{1}{3}(\frac{1}{(e^y+1)^3})=-\frac{1}{6}(\frac{1}{(e^x+1)^6}+C[/tex]
[tex]\frac{1}{3}(e^y+1)^{-3}=-\frac{1}{6}(e^x+1)^{-6}+C[/tex]
