Solve Separable Differential Equation (e^y + 1)^2e^(-y)dx + (e^x + 1)^3e^(-x)dy = 0

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Solve Separable Differential Equation (e^y + 1)^2e^(-y)dx + (e^x + 1)^2e^(-x)dy = 0

To solve this differential equation using the separable method, we need to write it in the form of f(y)dy=-g(x)dx where f(y) and g(x) are functions of only y and x, respectively. Starting with the given equation:

    \begin{equation*} (e^y+1)^2e^{-y}dx+(e^x+1)^3e^{-x}dy=0 \end{equation*}

We can divide both sides by (e^y+1)^2e^{-y}(e^x+1)^3 to get:

    \begin{equation*} \frac{dx}{(e^x+1)^3}+\frac{dy}{(e^y+1)^2}=0 \end{equation*}

Now we can integrate both sides with respect to their respective variables:

    \begin{equation*} \int \frac{dx}{(e^x+1)^3}+\int \frac{dy}{(e^y+1)^2}=C \end{equation*}

The integral on the left can be evaluated using substitution:

Let u=e^x+1, then du=e^xdx and the integral becomes:

    \begin{align*} \int \frac{dx}{(e^x+1)^3} &= \int \frac{du}{u^3} \ &= -\frac{1}{2u^2}+C_1 \ &= -\frac{1}{2(e^x+1)^2}+C_1 \end{align*}

Similarly, the integral on the right can be evaluated using substitution:

Let v=e^y+1, then dv=e^ydy and the integral becomes:

    \begin{align*} \int \frac{dy}{(e^y+1)^2} &= \int \frac{dv}{v^2} \ &= -\frac{1}{v}+C_2 \ &= -\frac{1}{e^y+1}+C_2 \end{align*}

Substituting these back into the original equation, we get:

    \begin{equation*} -\frac{1}{2(e^x+1)^2}-\frac{1}{e^y+1}=C \end{equation*}

where C=C_1+C_2.

Therefore, the solution to the differential equation is:

    \begin{equation*} \frac{1}{2(e^x+1)^2}+\frac{1}{e^y+1}=C \end{equation*}

where C is the constant of integration.

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