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dlinn [17]
3 years ago
13

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

Mathematics
1 answer:
Musya8 [376]3 years ago
3 0

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

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