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प्रश्न
(x2 − 2xy) dy + (x2 − 3xy + 2y2) dx = 0
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उत्तर
We have,
\[\left( x^2 - 2xy \right) dy + \left( x^2 - 3xy + 2 y^2 \right) dx = 0\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x^2 - 3xy + 2 y^2}{2xy - x^2}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x^2 - 3v x^2 + 2 v^2 x^2}{2v x^2 - x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1 - 3v + 2 v^2}{2v - 1}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - 3v + 2 v^2}{2v - 1} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 - 2v}{2v - 1}\]
\[ \Rightarrow x\frac{dv}{dx} = - 1\]
\[ \Rightarrow dv = - \frac{1}{x}dx\]
Integrating both sides, we get
\[\int dv = - \int\frac{1}{x}dx\]
\[ \Rightarrow v = - \log \left| x \right| + C\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \frac{y}{x} = - \log \left| x \right| + C\]
\[ \Rightarrow \frac{y}{x} + \log \left| x \right| = C\]
\[\text{ Hence, }\frac{y}{x} + \log \left| x \right| = C\text{ is the required solution .}\]
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