Exams Joint Entrance Examination Mathematics Differential Equations Question 466 Question 466 of 490 Medium MCQ
A function y = f ( x ) y=f(x) y = f ( x ) satisfies \[f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0\] with condition f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 . Then, \[f\left(\frac{\pi}{2}\right)\] is equal to
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\[ d y d x − ( sin 2 x 1 + cos 2 x ) y = sin x I.F. = 1 + cos 2 x y ⋅ ( 1 + cos 2 x ) = ∫ ( sin x ) d x = − cos x + C x = 0 , C = 1 y ( π 2 ) = 1 \begin{aligned}
& \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\sin x \\
& \text { I.F. }=1+\cos ^2 x \\
& y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \\
& =-\cos x+C \\
& x=0, C=1 \\
& y\left(\frac{\pi}{2}\right)=1
\end{aligned} d x d y − ( 1 + cos 2 x sin 2 x ) y = sin x I.F. = 1 + cos 2 x y ⋅ ( 1 + cos 2 x ) = ∫ ( sin x ) d x = − cos x + C x = 0 , C = 1 y ( 2 π ) = 1 \] Answer the question to unlock