Let g be a differentiable function such that ∫ g(t) dt t ∈ to (0, x) = x - ∫ tg (t) dt t ∈ to (0, x), x ≥ 0 and let y = y(x) satisfy

Question

Let g be a differentiable function such that \(\int_\limits{0}^{x} g(t) d t=x-\int_\limits{0}^{x} \operatorname{tg}(t) d t, x \geq 0\) and let y = y(x) satisfy the differential equation \(\frac{d y}{d x}-y \tan x=2(x+1) \sec x \ g(x), x \in\left[0, \frac{\pi}{2}\right).\) If y(0) = 0, then \(\mathrm{y}\left(\frac{\pi}{3}\right)\) is equal to

(1) \(\frac{2 \pi}{3 \sqrt{3}}\)

(2) \(\frac{4 \pi}{3}\)

(3) \(\frac{2 \pi}{3}\)

(4) \(\frac{4 \pi}{3 \sqrt{3}}\)

Answer 1

Correct option is: (2) \(\frac{4 \pi}{3}\)   

Diff. w.r.t. x

\(\mathrm{g}(\mathrm{x})=1-\mathrm{xg}(\mathrm{x})\)

\(g(x)=\frac{1}{1+x}\)

so \(\frac{d y}{d x}-y \tan x=2 \sec x\)

\(I F=e^{-\int \tan d x}=e^{\log \cos x}=\cos x\)

solution of D.E.

\(y \cos x=\int 2 d x+c\)

\(y \cos x=2 x+c\)

y(0) = 0

\(\mathrm{c}=0\)

\(y=\frac{2 x}{\cos x}\)

\(\mathrm{y}=2 \mathrm{x} \sec \mathrm{x}\)

\(\mathrm{y}\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2=\frac{4 \pi}{3}\)