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

Question

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

UpStudy AI · Accepted Answer

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
**General Solutions:** - **Problem (a):** $$ y(x) = -\frac{1}{2} \cos(2x) + C $$ - **Problem (b):** $$ y(x) = \frac{1}{2} + C e^{-2x} $$

Find the general solution of each of the following differential equati

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