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प्रश्न
The solution of the differential equation x dx + y dy = x2 y dy − y2 x dx, is
पर्याय
x2 − 1 = C (1 + y2)
x2 + 1 = C (1 − y2)
x3 − 1 = C (1 + y3)
x3 + 1 = C (1 − y3)
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उत्तर
x2 − 1 = C (1 + y2)
We have,
x dx + y dy = x2y dy − y2x dx
\[\Rightarrow \left( x + x y^2 \right)dx = \left( x^2 y - y \right)dy\]
\[ \Rightarrow \frac{x}{\left( x^2 - 1 \right)}dx = \frac{y}{\left( 1 + y^2 \right)}dy\]
\[ \Rightarrow \frac{2x}{2\left( x^2 - 1 \right)}dx = \frac{2y}{2\left( 1 + y^2 \right)}dy\]
Integrating both sides, we get
\[\frac{1}{2}\int\frac{2y}{\left( 1 + y^2 \right)}dy = \frac{1}{2}\int\frac{2x}{\left( x^2 - 1 \right)}dx\]
\[ \Rightarrow \frac{1}{2}\log\left| \left( 1 + y^2 \right) \right| = \frac{1}{2}\log\left| \left( x^2 - 1 \right) \right| - \frac{1}{2}\log\left| C \right|\]
\[ \Rightarrow \log\left| \left( 1 + y^2 \right) \right| = \log\left| \left( x^2 - 1 \right) \right| - \log\left| C \right|\]
\[ \Rightarrow \log\left| \left( 1 + y^2 \right) \right| = \log\left| \left( \frac{x^2 - 1}{C} \right) \right|\]
\[ \Rightarrow 1 + y^2 = \frac{x^2 - 1}{C}\]
\[ \Rightarrow C\left( 1 + y^2 \right) = x^2 - 1\]
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