Question

If $\frac{dy}{dx} + 2y \tan x = \sin x$ and $y = 0$, when $x = \frac{\pi}{3}$, show that the maximum value of $y$ is $\frac{1}{3}$.

Answer 1

The maximum value of y is calculated to be $\frac{1}{8}$. The question asks to show that the maximum value is $\frac{1}{3}$. Based on the derivation, the statement that the maximum value is $\frac{1}{3}$ is incorrect. The correct maximum value is $\frac{1}{8}$.

Answer 2

The given differential equation is $\frac{dy}{dx} + 2y \tan x = \sin x$. This is a first-order linear differential equation. The integrating factor (I.F.) is $e^{\int 2 \tan x dx} = e^{2 \ln |\sec x|} = \sec^2 x$. The general solution is $y \sec^2 x = \int \sec^2 x \sin x dx = \int \sec x \tan x dx = \sec x + C$. Applying the initial condition $y=0$ when $x = \frac{\pi}{3}$, we get $0 \cdot \sec^2\left(\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) + C$, which simplifies to $0 = 2 + C$, so $C = -2$. The specific solution is $y \sec^2 x = \sec x - 2$, or $y = \frac{\sec x - 2}{\sec^2 x} = \cos x - 2 \cos^2 x$. To find the maximum value of $y$, let $u = \cos x$. Then $y(u) = u - 2u^2$. This is a quadratic function whose maximum occurs at $u = -\frac{1}{2(-2)} = \frac{1}{4}$. The maximum value of $y$ is $y_{max} = \frac{1}{4} - 2\left(\frac{1}{4}\right)^2 = \frac{1}{4} - \frac{2}{16} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$. The question states that the maximum value is $\frac{1}{3}$, but the calculation shows it is $\frac{1}{8}$.