Question

cosec x (log y) dy + x²y dx = 0

Answer 1

We have,

\[cosec\ x \left( \log y \right)dy + x^2 y dx = 0\]

\[ \Rightarrow \frac{\log y}{y}dy = - \frac{x^2}{cosec\ x}dx\]

\[ \Rightarrow \frac{\log y}{y}dy = - x^2 \sin x dx\]

Integrating both sides, we get

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

Putting log y = t

\[\frac{1}{y}dy = dt\]

Therefore, (1) becomes

\[\int t\ dt = - \int x^2 \sin x dx\]

\[ \Rightarrow \frac{1}{2} \left[ \log y \right]^2 = - x^2 \int\sin x dx - \int\left( \frac{d}{dx}\left( x^2 \right)\int\sin x dx \right)dx\]

\[ \Rightarrow \frac{1}{2} \left[ \log y \right]^2 = x^2 \cos x + 2\int x \cos x dx\]

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

\[ \Rightarrow \frac{1}{2} \left[ \log y \right]^2 = x^2 \cos x - 2x \sin x - 2\cos x + C\]

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

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