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Question
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Solution
We have,
\[x\frac{dy}{dx} = x + y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + y}{x}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x + vx}{x}\]
\[ \Rightarrow v + x\frac{dv}{dx} = 1 + v\]
\[ \Rightarrow x\frac{dv}{dx} = 1 + v - v\]
\[ \Rightarrow x\frac{dv}{dx} = 1\]
\[ \Rightarrow dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int dv = \int\frac{1}{x}dx\]
\[ \Rightarrow v = \log \left| x \right| + C\]
\[\text{Putting }v = \frac{y}{x},\text{ we get}\]
\[ \Rightarrow \frac{y}{x} = \log \left| x \right| + C\]
\[ \Rightarrow y = x\log \left| x \right| + Cx \]
\[\text{ Hence, }y = x\log \left| x \right| + Cx\text{ is the required solution }.\]
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