Question

Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]

Answer 1

We have,

\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
\[ \Rightarrow \text{ cosec }x \log y \frac{dy}{dx} = - x^2 y^2 \]
\[ \Rightarrow \frac{1}{y^2}\log y dy = - \frac{x^2}{\text{ cosec }x}dx\]
\[ \Rightarrow \frac{1}{y^2}\log y dy = - x^2 \sin x dx\]
\[ \Rightarrow \int\frac{1}{y^2}\log y dy = - \int x^2 \sin x dx\]
\[\Rightarrow - \frac{\log y}{y} + \int\frac{1}{y} \times \frac{1}{y} = - \left[ - x^2 \cos x + \int2x\cos x dx \right] + C\]
\[ \Rightarrow - \frac{\log y}{y} - \frac{1}{y} = - \left[ - x^2 \cos x + 2x\sin x - 2\int\sin x dx \right] + C\]
\[ \Rightarrow - \left( \frac{1 + \log y}{y} \right) = - \left[ - x^2 \cos x + 2x\sin x + 2\cos x dx \right] + C\]
\[ \Rightarrow - \left( \frac{1 + \log y}{y} \right) - x^2 \cos x + 2\left( x\sin x + \cos x \right) = C\]