JEE Mains · Maths · STD 12 - 9. differential equations
A function \(y=f(x)\) satisfies \(f(x) \sin 2 x+\sin x-\left(1+\cos ^2 x\right) f^{\prime}(x)=0\) with condition \(f(0)=0\). Then \(f\left(\frac{\pi}{2}\right)\) is equal to
- A \(1\)
- B \(0\)
- C \(-1\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\( \frac{d y}{d x}-\left(\frac{\sin 2 x}{1+\cos ^2 x}\right) y=\left(\frac{\sin x}{1+\cos ^2 x}\right) \) \( \text { I.F. }=1+\cos ^2 x \) \( y \cdot\left(1+\cos ^2 x\right)=\int(\sin x) d x \) \( =-\cos x+C \) \( x=0, C=1 \) \( y\left(\frac{\pi}{2}\right)=1\)
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