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Mathematics Question on Differential equations

If
dydx+2ytan⁡⁡x=sin⁡⁡x,0<x<π2and y(π3)=0\frac{dy}{dx}+2y \tan⁡x=\sin⁡x, 0<x<\frac{π}{2} and\ y(\frac{π}{3})=0
then the maximum value of y(x) is:

A

18\frac{1}{8}

B

34\frac{3}{4}

C

14\frac{1}{4}

D

38\frac{3}{8}

Answer

18\frac{1}{8}

Explanation

Solution

dydx+2ytan⁡⁡x=sin⁡⁡x\frac{dy}{dx}+2y \tan⁡x=\sin⁡x
which is a first order linear differential equation.
Integrating factor (I.F.)=e∫2tan⁡x dx\text{Integrating factor (I.F.)} = e^{\int 2\tan x \,dx}
=e2In∣sec⁡⁡x∣=sec⁡2⁡=x=e^{2In|\sec ⁡x|}=\sec^2⁡ =x
Solution of differential equation can be written as
y⋅sec⁡2⁡x=∫sin⁡⁡x⋅sec⁡2⁡xdx=∫sec⁡⁡x⋅tan⁡⁡xdxy⋅\sec^2⁡x=∫\sin⁡x⋅\sec^2⁡x dx=∫\sec⁡x⋅\tan⁡x dx
ysec⁡2⁡x=sec⁡⁡x+Cy\sec^2⁡x=\sec⁡x+C
y(π3)=0,0=sec⁡⁡π3+Cy(\frac{π}{3})=0,0=\sec⁡ \frac{π}{3}+C
⇒C=−2
y=sec⁡⁡x−2sec⁡2⁡xy=\frac{\sec⁡x−2}{\sec^2⁡x}
y=cos⁡⁡x−2cos⁡2⁡xy=\cos⁡x−2\cos^2⁡x
=18−2(cos⁡⁡x−14)2=\frac{1}{8}−2(\cos⁡ x −\frac{1}{4})^2
ymax=18y_{max}=\frac{1}{8}
So, the correct option is (A): 18\frac{1}{8}