Step 1: Understanding the Concept:
This question asks us to classify a given first-order differential equation. We need to check if it fits the standard forms for variable separable, linear, homogeneous, or exact equations.
Step 2: Key Formula or Approach:
The standard forms for different types of first-order DEs are:
Variable Separable: Can be written as $f(y) dy = g(x) dx$.
Linear: Can be written as $\frac{dy}{dx} + P(x)y = Q(x)$.
Homogeneous: Can be written as $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.
Exact: Can be written as $M(x,y)dx + N(x,y)dy = 0$ where $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.
Step 3: Detailed Explanation:
The given equation is $\frac{dy}{dx} = -\frac{x+y}{1+x^2}$.
Let's rearrange the equation to test the standard forms.
\[ \frac{dy}{dx} = -\frac{x}{1+x^2} - \frac{y}{1+x^2} \]
Move the term with 'y' to the left side:
\[ \frac{dy}{dx} + \frac{1}{1+x^2} y = -\frac{x}{1+x^2} \]
This equation is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where:
\[ P(x) = \frac{1}{1+x^2} \quad \text{and} \quad Q(x) = -\frac{x}{1+x^2} \]
This is the standard form of a first-order linear differential equation.
Let's check why other options are incorrect:
- Variable Separable: The term $x+y$ in the numerator prevents separating x and y terms.
- Homogeneous: Let $f(x,y) = -\frac{x+y}{1+x^2}$. Then $f(\lambda x, \lambda y) = -\frac{\lambda x+\lambda y}{1+\lambda^2 x^2} = -\frac{\lambda(x+y)}{1+\lambda^2 x^2} \neq \lambda^0 f(x,y)$. So it is not homogeneous.
- Exact: Rewrite as $(x+y)dx + (1+x^2)dy = 0$. Here $M=x+y$ and $N=1+x^2$. $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 2x$. Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, it is not exact.
Step 4: Final Answer:
The equation fits the standard form of a first-order linear equation. Therefore, option (B) is correct.