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प्रश्न
Find the particular solution of the differential equation \[\left( x - y \right)\frac{dy}{dx} = x + 2y\], given that when x = 1, y = 0.
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उत्तर
\[\left( x - y \right)\frac{dy}{dx} = x + 2y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{x - y}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx \text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v - v + v^2}{1 - v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + v + v^2}{1 - v}\]
\[ \Rightarrow \frac{1 - v}{1 + v + v^2} dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{1 - v}{1 + v + v^2} dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{1}{1 + v + v^2}dv - \frac{1}{2}\int\frac{2v + 1 - 1}{1 + v + v^2} = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{1}{1 + v + v^2}dv - \frac{1}{2}\int\frac{2v + 1}{1 + v + v^2}dv + \frac{1}{2}\int\frac{1}{1 + v + v^2}dv = \int\frac{1}{x}dx \]
\[ \Rightarrow \frac{3}{2}\int\frac{1}{1 + v + v^2}dv - \frac{1}{2}\int\frac{2v + 1}{1 + v + v^2}dv = \int\frac{1}{x}dx \]
\[ \Rightarrow \frac{3}{2}\int\frac{1}{1 + v + v^2 + \frac{1}{4} - \frac{1}{4}}dv - \frac{1}{2}\int\frac{2v + 1}{1 + v + v^2}dv = \int\frac{1}{x}dx \]
\[ \Rightarrow \frac{3}{2}\int\frac{1}{\left( v + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}dv - \frac{1}{2}\int\frac{2v + 1}{1 + v + v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \sqrt{3}\tan {}^{- 1} \left| \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right| - \frac{1}{2}\log \left| 1 + v + v^2 \right| = \log \left| x \right| + C\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[\sqrt{3} \tan^{- 1} \left| \frac{2y + x}{\sqrt{3}x} \right| - \frac{1}{2}\log \left| \frac{x^2 + xy + y^2}{x^2} \right| = \log \left| x \right| + C\]
\[ \Rightarrow \sqrt{3} \tan^{- 1} \left| \frac{2y + x}{\sqrt{3}x} \right| - \frac{1}{2}\log \left| x^2 + xy + y^2 \right| + \log \left| x \right| = \log \left| x \right| + C \]
\[ \Rightarrow \sqrt{3} \tan^{- 1} \left| \frac{2y + x}{\sqrt{3}x} \right| - \frac{1}{2}\log \left| x^2 + xy + y^2 \right| = C . . . . . (1) \]
\[\text{ At }x = 1, y = 0 ...........\left(\text{Given} \right)\]
\[\text{ Putting }x = 1\text{ and }y = 0\text{ in }(1),\text{ we get }\]
\[\sqrt{3} \tan^{- 1} \left| \frac{1}{\sqrt{3}} \right| - \frac{1}{2}\log \left| 1 \right| = C\]
\[ \Rightarrow C = \sqrt{3} \tan^{- 1} \left| \frac{1}{\sqrt{3}} \right|\]
\[ \Rightarrow C = \sqrt{3} \times \frac{\pi}{6}\]
\[ \Rightarrow C = \frac{\pi}{2\sqrt{3}}\]
Substituting the value of C in (1), we get
\[\sqrt{3} \tan^{- 1} \left| \frac{2y + x}{\sqrt{3}x} \right| - \frac{1}{2}\log \left| x^2 + xy + y^2 \right| = \frac{\pi}{2\sqrt{3}}\]
\[ \Rightarrow 2\sqrt{3} \tan^{- 1} \left| \frac{2y + x}{\sqrt{3}x} \right| - \log \left| x^2 + xy + y^2 \right| = \frac{\pi}{\sqrt{3}}\]
\[ \Rightarrow \log \left| x^2 + xy + y^2 \right| = 2\sqrt{3} \tan^{- 1} \left| \frac{2y + x}{\sqrt{3}x} \right| - \frac{\pi}{\sqrt{3}}\]
\[\text{ Hence, }\log \left| x^2 + xy + y^2 \right| = 2\sqrt{3} \tan^{- 1} \left| \frac{2y + x}{\sqrt{3}x} \right| - \frac{\pi}{\sqrt{3}}\text{ is the required solution.}\]
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