JEE Mains · Maths · STD 12 - 9. differential equations
If \(\frac{ dy }{ dx }+2 y \tan x =\sin x , 0< x <\frac{\pi}{2}\) and \(y \left(\frac{\pi}{3}\right)=\) 0 , then the maximum value of \(y(x)\) is.
- A \(\frac{1}{8}\)
- B \(\frac{3}{4}\)
- C \(\frac{1}{4}\)
- D \(\frac{3}{8}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+2 y \tan x=\sin x\) \(I \cdot F=e^{\int 2 \tan x d x}=e^{\operatorname{mn}(\sec x)^{2}}=\sec ^{2} x\) \(y\left(\sec ^{2} x\right)=\int \sin x \sec ^{2} x d x+C\) \(y \cdot \sec ^{2} x=\sec x+C\) Put \(x=\frac{\pi}{3}, y=0\) \(y=\cos x-2 \cos ^{2} x\)…
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