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प्रश्न
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उत्तर
We have,
\[\cos x\frac{dy}{dx} - \cos 2x = \cos 3x\]
\[ \Rightarrow dy = \frac{\cos 3x + \cos 2x}{\cos x}dx\]
\[ \Rightarrow dy = \frac{4 \cos^3 x - 3\cos x + 2 \cos^2 x - 1}{\cos x}dx\]
\[ \Rightarrow dy = \left( 4 \cos^2 x - 3 + 2\cos x - \sec x \right)dx\]
\[ \Rightarrow dy = \left[ 2\left( 2 \cos^2 x - 1 \right) - 1 + 2\cos x - \sec x \right]dx\]
\[ \Rightarrow dy = \left( 2\cos 2x - 1 + 2\cos x - \sec x \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( 2\cos 2x - 1 + 2\cos x - \sec x \right)dx\]
\[ \Rightarrow y = \sin 2x - x + 2\sin x - \log\left| \sec x + \tan x \right| + C\]
\[\text{ Hence, } y = \sin 2x - x + 2\sin x - \log\left| \sec x + \tan x \right| +\text{C is the solution to the given differential equation.}\]
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