\(\frac{dy}{dx}+\frac{3}{cos^2x}y=\frac1{cos^2x}\)
I. F. = e∫ρdx = e∫\(\frac{3}{cos^2x}dx\) = e3∫\(sec^2x\) dx
= e3tanx
Complete solution is
y x I.F = ∫(I. F) x Q dx
⇒ y e3tan x = ∫sec2x e3tanxdx
= ∫e3t df (By taking t = tan x, \(\therefore\) df = sec2x dx)
⇒ y e3tanx = \(\frac{e^{3t}}{3}+c\) (By putting t = tan x)
\(\because\) y(π/4) = 4/3
\(\therefore\) 4/3 x e3tan π/4 = \(\frac{e^{3 tan\pi/4}}3+c\)
⇒ c = 4/3 e3 - e3/3 = e3
\(\therefore\) y e3tan x = \(\frac{e^{3 tan\,x}}3+e^3\)
⇒ y = 1/3 + e3 e-3 tan x = 1/3 + e3(1- tan x)
Now, y (-π/4) = 1/3 + e3(1-tan(-π/4)) = 1/3 + e3(1 + 1)
= 1/3 + e6
option (A) is correct.