Question:

The given differential equation is $\frac{dy}{dx} - 2y\cot 2x = \cos 2x$. This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -2\cot 2x$ and $Q(x) = \cos 2x$. Step 1: Determine the integrating factor. The integrating factor (IF) is given by $e^{\int P(x) dx}$. $$ \int P(x) dx = \int -2\cot 2x dx $$ To evaluate this integral, let $u = \sin 2x$, so $du = 2\cos 2x dx$. Then $\cot 2x dx = \frac{\cos 2x}{\sin 2x} dx = \frac{1}{u} \frac{du}{2}$. $$ \int -2\cot 2x dx = -2 \int \frac{1}{u} \frac{du}{2} = -\int \frac{1}{u} du = -\ln|u| = -\ln|\sin 2x| $$ Thus, the integrating factor is: $$ \text{IF} = e^{-\ln|\sin 2x|} = e^{\ln(|\sin 2x|^{-1})} = \frac{1}{|\sin 2x|} $$ For $x$ in an interval where $\sin 2x > 0$ (e.g., near $\pi/4$ or $0^+$), we use $\text{IF} = \frac{1}{\sin 2x} = \csc 2x$. Step 2: Solve the differential equation. Multiply the differential equation by the integrating factor: $$ \frac{1}{\sin 2x} \frac{dy}{dx} - \frac{2y\cot 2x}{\sin 2x} = \frac{\cos 2x}{\sin 2x} $$ The left side is the derivative of $(y \cdot \text{IF})$: $$ \frac{d}{dx}\left(y \cdot \frac{1}{\sin 2x}\right) = \frac{\cos 2x}{\sin 2x} = \cot 2x $$ Integrate both sides with respect to $x$: $$ y \cdot \frac{1}{\sin 2x} = \int \cot 2x dx $$ Using the substitution $u = \sin 2x$, $du = 2\cos 2x dx$: $$ \int \cot 2x dx = \int \frac{\cos 2x}{\sin 2x} dx = \int \frac{1}{u} \frac{du}{2} = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln|\sin 2x| + C $$ So, the general solution is: $$ \frac{y}{\sin 2x} = \frac{1}{2}\ln|\sin 2x| + C $$ $$ y(x) = \frac{\sin 2x}{2}\ln|\sin 2x| + C\sin 2x $$ Step 3: Apply the initial condition $y(\pi/4) = 0$. Substitute $x = \pi/4$ into the general solution: $$ y(\pi/4) = \frac{\sin(2 \cdot \pi/4)}{2}\ln|\sin(2 \cdot \pi/4)| + C\sin(2 \cdot \pi/4) $$ $$ 0 = \frac{\sin(\pi/2)}{2}\ln|\sin(\pi/2)| + C\sin(\pi/2) $$ Since $\sin(\pi/2) = 1$ and $\ln(1) = 0$: $$ 0 = \frac{1}{2}(0) + C(1) $$ $$ 0 = C $$ The particular solution is therefore: $$ y(x) = \frac{\sin 2x}{2}\ln|\sin 2x| $$ Step 4: Find $y(0)$. The function $y(x)$ is not directly defined at $x=0$ because $\sin 2x = 0$ and $\ln(0)$ is undefined. We evaluate the limit as $x \to 0$: $$ y(0) = \lim_{x \to 0} \left(\frac{\sin 2x}{2}\ln|\sin 2x|\right) $$ Let $u = \sin 2x$. As $x \to 0^+$, $u \to 0^+$. The limit becomes: $$ \lim_{u \to 0^+} \left(\frac{u}{2}\ln u\right) $$ This is an indeterminate form of type $0 \cdot (-\infty)$. We rewrite it as $\frac{\ln u}{2/u}$ to apply L'Hôpital's Rule: $$ \lim_{u \to 0^+} \frac{\ln u}{2/u} = \lim_{u \to 0^+} \frac{1/u}{-2/u^2} = \lim_{u \to 0^+} \frac{1}{u} \cdot \left(-\frac{u^2}{2}\right) = \lim_{u \to 0^+} \left(-\frac{u}{2}\right) = 0 $$ Therefore, $y(0) = 0$.