Question

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

• A. $\frac{1}{3}+e^6$
• B. $\frac{1}{3}$
• C. $-\frac{4}{3}$
• D. $\frac{1}{3}+e^3$

Answer 1

Answer: (A)

$\frac{1}{3}+e^6$

$\frac{dy}{dx}+3{\sec }^{2}x\cdot y={\sec }^{2}x$
I.F. $=e^{3\int {\sec }^{2}xdx}=e^{3\tan x}$

or $y\cdot e^{3\tan x}=\int {\sec }^{2}x\cdot e^{3\tan x}dx$
or $y\cdot e^{3\tan x}=\frac{1}{3}{e}^{3\tan x}+C$ .$\left(1\right)$

Given
$y\left(\frac{\pi }{4}\right)=\frac{4}{3}$
$\therefore \frac{4}{3}\cdot e^3=\frac{1}{3}{e}^3+C$

$\therefore C=e^3$

Now put $x=-\frac{\pi }{4}$ in equation $\left(1\right)$
$\therefore y\cdot e^{-3}=\frac{1}{3}{e}^{-3}+e^3$

$\therefore y=\frac{1}{3}+e^6$

$\therefore y(-\frac{\pi }{4})=\frac{1}{3}+e^6$.