Question

If $y=y(x)$ is the solution of the differential equation, $\frac{dy}{dx}+2y\tan x=\sin x,y\left(\frac{\pi }{3}\right)=0$, then the maximum value of the function $y(x)$ over $R$ is equal to :

• A. $8$
• B. $\frac{1}{2}$
• C. $-\frac{15}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (D)

$\frac{dy}{dx}+2y\tan x=\sin x$
$I\cdot F_1=e^{\int 2\tan xdx}=e^{2t\ln \sec x}$
$I\cdot F_{.}={\sec }^{2}x$
$y\cdot \left({\sec }^{2}x\right)=\int \sin x\cdot {\sec }^{2}xdx$
$y\cdot \left({\sec }^{2}x\right)=\int \sec x\tan xdx$
$y\cdot \left({\sec }^{2}x\right)=\sec x+C$
$x=\frac{\pi }{3};y=0$
$\Rightarrow C=-2$
$\Rightarrow y=\frac{\sec x-2}{{\sec }^{2}x}=\cos x-2{\cos }^{2}x$
$y=t-2t^2$
$\Rightarrow \frac{dy}{dt}=1-4t=0$
$\Rightarrow t=\frac{1}{4}$
$\therefore \max =\frac{1}{4}-\frac{1}{8}=\frac{2-1}{8}=\frac{1}{8}$