Which of the following equation(s) is/are linear?

Options

A. `(dy)/(dx)+y/x=logx`
B. `y(dy)/(dx)+4x=0`
C. `(2x+y^(3))(dy)/(dx)=3y`
D. None of these

Explanation

To determine which of the given equations are linear differential equations, we will analyze each equation based on the standard form of a linear differential equation, which is given by:

\[ \frac{dy}{dx} + p(x)y = q(x) \]

where \(p(x)\) and \(q(x)\) are functions of \(x\).

### Step-by-step Solution:

1. **Examine the First Equation:**
\[ \frac{dy}{dx} + \frac{y}{x} = \log(x) \]
- Here, we can identify \(p(x) = \frac{1}{x}\) and \(q(x) = \log(x)\).
- This equation is in the standard linear form, so **the first equation is linear**.

2. **Examine the Second Equation:**
\[ y \frac{dy}{dx} + 4x = 0 \]
- Rearranging gives:
\[ y \frac{dy}{dx} = -4x \]
- Dividing both sides by \(y\) (assuming \(y \neq 0\)):
\[ \frac{dy}{dx} = -\frac{4x}{y} \]
- This is not in the form \(\frac{dy}{dx} + p(x)y = q(x)\) because of the \(y\) in the denominator. Thus, **the second equation is not linear**.

3. **Examine the Third Equation:**
\[ 2x + y^3 \frac{dy}{dx} = 3y \]
- Rearranging gives:
\[ y^3 \frac{dy}{dx} = 3y - 2x \]
- Dividing both sides by \(y^3\) (assuming \(y \neq 0\)):
\[ \frac{dy}{dx} = \frac{3y - 2x}{y^3} \]
- This can be rewritten as:
\[ \frac{dy}{dx} - \frac{2x}{y^3} = \frac{3}{y^2} \]
- Here, we can identify \(p(y) = -\frac{2}{y^3}\) and \(q(y) = \frac{3}{y^2}\). This is in the linear form with respect to \(y\). Therefore, **the third equation is linear**.

### Conclusion:
- The first and third equations are linear differential equations.
- The second equation is not linear.

### Summary of Results:
- **Linear Equations:** 1 and 3
- **Non-linear Equation:** 2