JEE Mains · Maths · STD 12 - 9. differential equations
If \( y=y(x) \) satisfies the differential equation \( 16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})cos~y~dy=(1+2~sin~y)dx, x>0 \) and \( y(256)=\frac{\pi}{2}, y(49)=\alpha \) then \( 2~sin~\alpha \) is equal to:
- A \(2 \sqrt{2}-1\)
- B \(2(\sqrt{2}-1)\)
- C \(3(\sqrt{2}-1)\)
- D \(\sqrt{2}-1\)
Answer & Solution
Correct Answer
(A) \(2 \sqrt{2}-1\)
Step-by-step Solution
Detailed explanation
\(\int \frac{\cos y}{1+2 \sin y} d y=\int \frac{d x}{16(\sqrt{9 \sqrt{x}+x})(4+\sqrt{9+\sqrt{x}})}\) \(4+\sqrt{9+\sqrt{x}}=t\) \(\frac{1}{2 \sqrt{9+\sqrt{x}}} \times \frac{d x}{2 \sqrt{x}}=1 d x\) \(\frac{1}{2} \ell n |1+2 \sin y |=\int \frac{4 dt }{16 t }+ C\)…
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