Question

The solution of the differential equation x dx + y dy = x² y dy − y² x dx, is

विकल्प

• x² − 1 = C (1 + y²)

• x² + 1 = C (1 − y²)

• x³ − 1 = C (1 + y³)

• x³ + 1 = C (1 − y³)

Answer 1

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

\[\frac{1}{2}\int\frac{2y}{\left( 1 + y^2 \right)}dy = \frac{1}{2}\int\frac{2x}{\left( x^2 - 1 \right)}dx\]

\[ \Rightarrow \frac{1}{2}\log\left| \left( 1 + y^2 \right) \right| = \frac{1}{2}\log\left| \left( x^2 - 1 \right) \right| - \frac{1}{2}\log\left| C \right|\]

\[ \Rightarrow \log\left| \left( 1 + y^2 \right) \right| = \log\left| \left( x^2 - 1 \right) \right| - \log\left| C \right|\]

\[ \Rightarrow \log\left| \left( 1 + y^2 \right) \right| = \log\left| \left( \frac{x^2 - 1}{C} \right) \right|\]

\[ \Rightarrow 1 + y^2 = \frac{x^2 - 1}{C}\]

\[ \Rightarrow C\left( 1 + y^2 \right) = x^2 - 1\]