Let $g$ be a differentiable function such that $\int_{0}^{x}g(t)dt=x-\int_{0}^{x}tg(t)dt$ for $x \ge 0$. Let $y=y(x)$ satisfy the differential equation $\frac{dy}{dx}-y \tan x = 2(x+1)\sec x \cdot g(x)$ for $x \in [0, \frac{\pi}{2})$. If $y(0)=0$, then $y\left(\frac{\pi}{3}\right)$ is equal to
- (1) $\frac{2\pi}{3\sqrt{3}}$
- (2) $\frac{4\pi}{3}$
- (3) $\frac{2\pi}{3}$
- (4) $\frac{4\pi}{3\sqrt{3}}$
Solution:
$$ \int_{0}^{x} g(t) dt = x – \int_{0}^{x} t g(t) dt $$
$$ g(x) = 1 – x g(x) $$
$$ g(x) + x g(x) = 1 $$
$$ g(x)(1+x) = 1 \Rightarrow g(x) = \frac{1}{1+x} $$
$$ \frac{dy}{dx} – y \tan x = 2(x+1)\sec x \cdot g(x) $$
$$ \frac{dy}{dx} – y \tan x = 2(x+1)\sec x \cdot \left(\frac{1}{1+x}\right) $$
$$ \frac{dy}{dx} – y \tan x = 2\sec x $$
$$ \text{I.F.} = e^{\int -\tan x dx} = e^{\ln(\cos x)} = \cos x $$
$$ y \cdot (\cos x) = \int (2\sec x \cdot \cos x) dx + c $$
$$ y \cos x = \int 2 dx + c $$
$$ y \cos x = 2x + c $$
$$ y(0) = 0 \Rightarrow 0 \cdot \cos(0) = 2(0) + c \Rightarrow c = 0 $$
$$ y \cos x = 2x $$
$$ y = \frac{2x}{\cos x} = 2x \sec x $$
$$ y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right) \sec\left(\frac{\pi}{3}\right) $$
$$ y\left(\frac{\pi}{3}\right) = \frac{2\pi}{3} (2) = \frac{4\pi}{3} $$
Ans. (2)