JEE Mains · Maths · STD 12 - 9. differential equations
If \(y=y(x)\) is the solution of the differential equation, \(\frac{ dy }{ dx }+2 y \tan x =\sin x , y \left(\frac{\pi}{3}\right)=0,\) then the maximum value of the function \(y ( x )\) over \(R\) is equal to
- A \(8\)
- B \(\frac{1}{2}\)
- C \(-\frac{15}{4}\)
- D \(\frac{1}{8}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+2 y \tan x=\sin x\) \(I.F.\) \(=e^{\int 2 \text { tan } x d x}=e^{2 log \sec x}\) \(I.F.\) \(=\sec ^{2} x\) \(y.\) \(\left(\sec ^{2} x\right)=\int \sin x \cdot \sec ^{2} x d x\) \(y.\) \(\left(\sec ^{2} x\right)=\int \sec x \tan x d x\) \(y.\)…
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