Question

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

• A. \(\frac{1}{8}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{1}{4}\)
• D. \(\frac{3}{8}\)

Answer 1

The Correct Option is A

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