The general solution of the differential equation \( \displaystyle\frac{dy}{dx}+y\cot x=2\cos x \) is (Where C is an arbitrary constant)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (y-\sin x)\sin x=C \)
The equation is linear in y, with \( P=\cot x \) and \( Q=2\cos x \).
\( \displaystyle \text{I.F.}=e^{\int\cot x\,dx} \)
\( \displaystyle =e^{\log\sin x}=\sin x \)
Multiplying by the integrating factor and integrating:
\( \displaystyle y\sin x=\int 2\cos x\sin x\,dx \)
\( \displaystyle =\int\sin 2x\,dx \)
\( \displaystyle =-\frac{\cos 2x}{2}+C_1 \)
Since \( -\frac{\cos 2x}{2}=\frac{2\sin^2x-1}{2}=\sin^2x-\frac{1}{2} \), the constant can absorb the \( -\frac{1}{2} \):
\( \displaystyle y\sin x=\sin^2x+C \)
\( \displaystyle y\sin x-\sin^2x=C \)
\( \displaystyle (y-\sin x)\sin x=C \)