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

Answer 1

The correct option is B $\frac{1}{3}+e^6$

$\frac{dy}{dx}+\frac{3}{{\cos }^{2}x}y=\frac{1}{{\cos }^{2}x}$
$\Rightarrow \frac{dy}{dx}+3{\sec }^{2}x.y={\sec }^{2}x$
$\Rightarrow \int \frac{dy}{1-3y}=\int {\sec }^{2}xdx$
$\Rightarrow \frac{\ln |(1-3y)|}{-3}=\tan x+c\cdots \left(1\right)$

At $x=\frac{\pi }{4},y=\frac{4}{3}$
$\Rightarrow \frac{\ln |1-3\cdot \frac{4}{3}|}{-3}=\tan \frac{\pi }{4}+c$
$\Rightarrow \frac{\ln 3}{-3}=1+c$
$c=-1-\frac{1}{3}\ln 3$

From eqn $\left(1\right)$
$\Rightarrow \frac{\ln |1-3y|}{-3}=\tan x-1-\frac{1}{3}\ln 3$
At $x=\frac{-\pi }{4}$
$\ln |1-3y|=6+\ln 3$
$\Rightarrow \ln \left|\frac{1-3y}{3}\right|=6$
$\Rightarrow \left|\frac{1-3y}{3}\right|=e^6$
$\Rightarrow \frac{1-3y}{3}=\pm e^6$
$\Rightarrow y=\frac{1}{3}\mp e^6$