JEE Mains · Maths · STD 12 - 9. differential equations
If \(\frac{{dy}}{{dx}} + \frac{3}{{{{\cos }^2}\,x}}\,y = \frac{1}{{{{\cos }^2}\,x}},\) \(x \in \left( {\frac{{ - \pi }}{3},\frac{\pi }{3}} \right)\) and \(y\left( {\frac{\pi }{4}} \right) = \frac{4}{3}\), then \(y\left( { - \frac{\pi }{4}} \right)\) equals
- A \(\frac{1}{3} + {e^6}\)
- B \( \frac{1}{3}\)
- C \( - \frac{4}{3}\)
- D \(\frac{1}{3} + {e^3}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3} + {e^6}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+\left(3 \sec ^{2} x\right) y=\sec ^{2} x\) This is linear differential equation Integrating factor \(=e^{\int 3 \sec ^{2} x d x}=e^{3 \tan x}\) Hence \(y \cdot e^{3 \tan x}=e^{\int 3 \tan x} \cdot \sec ^{2} x d x\)…
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