Question

If $y=y\left(x\right)$ 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 $y\left(x\right)$ over $R$ is equal to:

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

Answer 1

The correct option is D $\frac{1}{8}$

$\frac{dy}{dx}+2\tan x.y=\sin xI.F.=e^{\ln \left({\sec }^{2}x\right)}={\sec }^{2}x\Rightarrow y{\sec }^{2}x=\int \tan x\sec xdx=\sec x+c\text{Now}x=\frac{\pi }{3},y=0\Rightarrow c=-2\therefore y=\cos x-2{\cos }^{2}xy=-2({\cos }^{2}x-\frac{1}{2}\cos x)=-2({(\cos x-\frac{1}{4})}^2-\frac{1}{16})\Rightarrow y=\frac{1}{8}-2{(\cos x-\frac{1}{4})}^2\therefore y_{max}=\frac{1}{8}$