Step 1: Understanding the Concept:
A first-order differential equation is called linear if it can be expressed in the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\) or \(\frac{dx}{dy} + P(y)x = Q(y)\). In this form, the dependent variable (y in the first case, x in the second) and its derivative appear only to the first power and are not multiplied together.
Step 3: Detailed Explanation:
Let's analyze each equation:
(A) $\frac{dy{dx} + P(x)y = Q(x)$:} This is the very definition of a linear first-order differential equation with y as the dependent variable. So, (A) is linear.
(B) $\frac{dx{dy} + P(y)x = Q(y)$:} This is the standard form of a linear first-order differential equation with x as the dependent variable. So, (B) is linear.
(C) $(x - y)\frac{dy{dx} = x + 2y$:} We can rewrite this as \(\frac{dy}{dx} = \frac{x+2y}{x-y}\). This equation cannot be arranged into either of the standard linear forms. The terms involve products of y and \(\frac{dy}{dx}\), and it's a homogeneous equation, not linear.
(D) $(1 + x^2)\frac{dy{dx} + 2xy = 2$:} To check if this is linear, we try to put it in the standard form. Divide the entire equation by \((1 + x^2)\):
\[ \frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{2}{1 + x^2} \]
This equation is exactly in the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = \frac{2x}{1 + x^2}\) and \(Q(x) = \frac{2}{1 + x^2}\). Therefore, (D) is a linear differential equation.
Step 4: Final Answer:
The equations (A), (B), and (D) are linear first-order differential equations. This corresponds to option (1).