Question

Find the particular solution of the differential equation \[ \frac{dy}{dx} - y \cot x = \sin 2x, \quad \text{given that } y = 2 \text{ when } x = \frac{\pi}{2}. \]

Hint:

To solve linear differential equations of the form \[ \frac{dy}{dx} + P(x)y = Q(x) \] use the integrating factor method: \begin{itemize} \item Find \( \text{I.F.} = e^{\int P(x) dx} \) \item Multiply the entire equation by the I.F. \item The LHS becomes a product derivative. \item Integrate both sides, then apply initial conditions if given. \end{itemize}

Answer 1

We are given the differential equation: \[ \frac{dy}{dx} - y \cot x = \sin 2x \] Step 1: Convert to standard linear form The standard form of a first-order linear differential equation is: \[ \frac{dy}{dx} + P(x)y = Q(x) \] Here, we rewrite: \[ \frac{dy}{dx} + (-\cot x) y = \sin 2x \] So, \( P(x) = -\cot x \), and \( Q(x) = \sin 2x \). Step 2: Find the Integrating Factor (I.F.) \[ \text{I.F.} = e^{\int P(x)\, dx} = e^{\int -\cot x\, dx} \] \[ = e^{-\ln|\sin x|} = \frac{1}{\sin x} \] Step 3: Multiply the equation by the I.F. Multiplying both sides by \( \frac{1}{\sin x} \): \[ \frac{1}{\sin x} \cdot \frac{dy}{dx} - \frac{y \cot x}{\sin x} = \frac{\sin 2x}{\sin x} \] We know \( \sin 2x = 2 \sin x \cos x \), so: \[ \frac{d}{dx} \left( \frac{y}{\sin x} \right) = 2 \cos x \] Step 4: Integrate both sides \[ \int \frac{d}{dx} \left( \frac{y}{\sin x} \right) dx = \int 2 \cos x \, dx \] \[ \Rightarrow \frac{y}{\sin x} = 2 \sin x + C \] Step 5: Solve for \( y \) \[ y = \sin x (2 \sin x + C) = 2 \sin^2 x + C \sin x \] Step 6: Apply the initial condition We are given: \( y = 2 \) when \( x = \frac{\pi}{2} \). Since \( \sin\left(\frac{\pi}{2}\right) = 1 \), we substitute: \[ 2 = 2(1)^2 + C(1) \Rightarrow 2 = 2 + C \Rightarrow C = 0 \] Final Answer: \[ \boxed{y = 2 \sin^2 x} \]