Question

If $y = y ( x )$ is the solution of the differential equation, $\frac{ dy }{ d x }+2 y \tan x =\sin x , y \left(\frac{\pi}{3}\right)=0$, 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)

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