Question

If $\frac{dy}{dx}+\frac{3}{{cos}^{2}x}=\frac{1}{{cos}^{2}x},xϵ(\frac{-\pi }{3},\frac{\pi }{3}),$and $y\left(\frac{\pi }{4}\right)=\frac{4}{3},$ then $y(-\frac{\pi }{4})$equals:

• A. $\frac{16}{3}$
• B. $\frac{1}{3}$
• C. $\frac{-4}{3}$
• D. $\frac{1}{3}+e^3$

Answer 1

The correct option is A $\frac{16}{3}$

$\frac{dy}{dx}+\frac{3}{co{s}^{2}x}=\frac{1}{co{s}^{2}x}$

$\frac{dy}{dx}=\frac{1}{co{s}^{2}x}-\frac{3}{co{s}^{2}x}$

$\frac{dy}{dx}=se{c}^{2}x-3se{c}^{2}x$

$\frac{dy}{dx}=-2se{c}^{2}x$

$\therefore \int dy=\int -2se{c}^{2}x0dx$ (Integerating)

$\therefore y=-2tanx+C$

Now $y\left(\frac{\pi }{4}\right)=\frac{4}{3}$

$\frac{4}{3}=-2tan\left(\frac{\pi }{4}\right)+C$

$\frac{4}{3}=-2+C$ $\therefore C=\frac{10}{3}$

$\therefore A+x=-\frac{\pi }{4}$

$y=-2tan(-\frac{\pi }{4})+\frac{10}{3}$

$=-2(-1)+\frac{10}{3}$

$=\frac{16}{3}$