Find the general solution of each of the following differential equati \( \begin{array}{ll}\text { (a) } \sec ^{2} x \frac{\mathrm{~d} y}{\mathrm{~d} x}=2 \tan x & \text { (b) } \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=1\end{array} \)

Let's solve each differential equation step-by-step. ### Problem (a) **Equation:** \[ \sec^{2}x \frac{dy}{dx} = 2 \tan x \] **Solution:** 1. **Rearrange the equation to solve for \(\frac{dy}{dx}\):** \[ \frac{dy}{dx} = \frac{2 \tan x}{\sec^2 x} \] 2. **Simplify using trigonometric identities:** \[ \tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x} \] \[ \frac{dy}{dx} = 2 \cdot \frac{\sin x}{\cos x} \cdot \cos^2 x = 2 \sin x \cos x \] 3. **Recognize a double-angle identity:** \[ 2 \sin x \cos x = \sin(2x) \] So, \[ \frac{dy}{dx} = \sin(2x) \] 4. **Integrate both sides with respect to \(x\):** \[ y = \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \] Where \(C\) is the constant of integration. **General Solution:** \[ y(x) = -\frac{1}{2} \cos(2x) + C \] --- ### Problem (b) **Equation:** \[ \frac{dy}{dx} + 2y = 1 \] **Solution:** This is a linear first-order differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where \( P(x) = 2 \) and \( Q(x) = 1 \). 1. **Find the integrating factor (\(\mu(x)\)):** \[ \mu(x) = e^{\int P(x) \, dx} = e^{2x} \] 2. **Multiply the entire differential equation by the integrating factor:** \[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x} \] 3. **Recognize the left-hand side as the derivative of \( e^{2x} y \):** \[ \frac{d}{dx} \left( e^{2x} y \right) = e^{2x} \] 4. **Integrate both sides with respect to \(x\):** \[ e^{2x} y = \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C \] 5. **Solve for \(y\):** \[ y = \frac{1}{2} + Ce^{-2x} \] Where \(C\) is the constant of integration. **General Solution:** \[ y(x) = \frac{1}{2} + C e^{-2x} \] Answer: Problem b Answer After simplifying, all solutions are equal to one half plus a constant times e^–2x. Thus y = ½ + C e^−2x