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The solution of the differential equation $\frac{d y}{d x}=\frac{x y+y}{x y+x}$ is
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Verified Answer
The correct answer is:
$y-x=\log \left(\frac{c x}{y}\right)$
Given differential equation is
$$
\begin{aligned}
\frac{d y}{d x} & =\frac{x y+y}{x y+x} \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{y(1+x)}{x(1+y)} \\
\Rightarrow \quad \frac{(1+y)}{y} d y & =\frac{(1+x)}{x} d x \\
\Rightarrow \quad \int\left(\frac{1}{y}+1\right) d y & =\int\left(\frac{1}{x}+1\right) d x \\
\Rightarrow \quad \log y+y & =\log x+x+\log c \\
\Rightarrow \quad y-x & =\log \left(\frac{c x}{y}\right)
\end{aligned}
$$
$$
\begin{aligned}
\frac{d y}{d x} & =\frac{x y+y}{x y+x} \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{y(1+x)}{x(1+y)} \\
\Rightarrow \quad \frac{(1+y)}{y} d y & =\frac{(1+x)}{x} d x \\
\Rightarrow \quad \int\left(\frac{1}{y}+1\right) d y & =\int\left(\frac{1}{x}+1\right) d x \\
\Rightarrow \quad \log y+y & =\log x+x+\log c \\
\Rightarrow \quad y-x & =\log \left(\frac{c x}{y}\right)
\end{aligned}
$$
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