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प्रश्न
The solution of the differential equation \[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}\], is
पर्याय
\[\tan^{- 1} \left( \frac{x}{y} \right) = \log y + C\]
\[\tan^{- 1} \left( \frac{y}{x} \right) = \log x + C\]
\[\tan^{- 1} \left( \frac{x}{y} \right) = \log x + C\]
\[\tan^{- 1} \left( \frac{y}{x} \right) = \log y + C\]
MCQ
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उत्तर
\[\tan^{- 1} \left( \frac{y}{x} \right) = \log x + C\]
We have,
\[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} \] ................(1)
This is homogenous differential equation.
\[\text{ Let }y = vx\]
\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{ Now, putting }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ and }y = vx\text{ in }\left( 1 \right),\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x^2 + x^2 v + x^2 v^2}{x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = 1 + v + v^2 \]
\[ \Rightarrow x\frac{dv}{dx} = 1 + v^2 \]
\[ \Rightarrow \left( \frac{1}{1 + v^2} \right)dv = \frac{1}{x}dx\]
Integrating both sides we get,
\[\int\frac{1}{1 + v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \tan^{- 1} v = \log x + C\]
\[ \Rightarrow \tan^{- 1} \left( \frac{y}{x} \right) = \log x + C\]
This is homogenous differential equation.
\[\text{ Let }y = vx\]
\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{ Now, putting }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ and }y = vx\text{ in }\left( 1 \right),\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x^2 + x^2 v + x^2 v^2}{x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = 1 + v + v^2 \]
\[ \Rightarrow x\frac{dv}{dx} = 1 + v^2 \]
\[ \Rightarrow \left( \frac{1}{1 + v^2} \right)dv = \frac{1}{x}dx\]
Integrating both sides we get,
\[\int\frac{1}{1 + v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \tan^{- 1} v = \log x + C\]
\[ \Rightarrow \tan^{- 1} \left( \frac{y}{x} \right) = \log x + C\]
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