Question

Which of the following equations is NOT a Linear Differential Equation?

• A. \((1 + x^2) \, dy + 2xy \, dx = \cot x \, dx\)
• B. \(y + \frac{d}{dx}(xy) = x(\sin x + \log x)\)
• C. \(x(1 + y^2) \, dx - y(1 + x^2) \, dy = 0\)
• D. \(y \, dx - (x + 3y^2) \, dy = 0\)

Hint:

To quickly check for linearity, look for powers of the dependent variable or its derivatives. If you spot terms like \(y^2\), \(\sin(y)\), or \(e^y\), and the equation cannot be rearranged to treat \(x\) as a linear dependent variable, it is non-linear.

Answer 1

The Correct Option is C

Concept: A differential equation is classified as a Linear Differential Equation if the dependent variable (typically \(y\)) and all of its derivatives occur only to the first power, and are never multiplied together. The standard form for a first-order linear differential equation is: \[ \frac{dy}{dx} + P(x)y = Q(x) \quad \text{or} \quad \frac{dx}{dy} + P_1(y)x = Q_1(y) \] If an equation contains terms like \(y^2\), \(\left(\frac{dy}{dx}\right)^2\), or \(y \frac{dy}{dx}\), it is non-linear.

Step 1: Evaluate Option (A)
\[ (1 + x^2) \, dy + 2xy \, dx = \cot x \, dx \] Dividing the entire equation by \(dx\): \[ (1 + x^2)\frac{dy}{dx} + 2xy = \cot x \] Dividing by \((1+x^2)\): \[ \frac{dy}{dx} + \left(\frac{2x}{1+x^2}\right)y = \frac{\cot x}{1+x^2} \] This fits the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\). Thus, it is linear.

Step 2: Evaluate Option (B)
\[ y + \frac{d}{dx}(xy) = x(\sin x + \log x) \] Applying the product rule to expand the derivative term: \(\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\). \[ y + x\frac{dy}{dx} + y = x(\sin x + \log x) \quad \Rightarrow \quad x\frac{dy}{dx} + 2y = x(\sin x + \log x) \] Dividing by \(x\): \[ \frac{dy}{dx} + \left(\frac{2}{x}\right)y = \sin x + \log x \] This fits the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\). Thus, it is linear.

Step 3: Evaluate Option (C)
\[ x(1 + y^2) \, dx - y(1 + x^2) \, dy = 0 \] Rearranging terms to find an expression for \(\frac{dy}{dx}\): \[ y(1 + x^2) \, dy = x(1 + y^2) \, dx \quad \Rightarrow \quad \frac{dy}{dx} = \frac{x(1+y^2)}{y(1+x^2)} \] Notice the presence of the \(y^2\) term. Because the dependent variable \(y\) appears with an exponent of $2$, this equation cannot be written in a linear form. Thus, it is not linear.

Step 4: Evaluate Option (D)
\[ y \, dx - (x + 3y^2) \, dy = 0 \quad \Rightarrow \quad y \frac{dx}{dy} - x = 3y^2 \quad \Rightarrow \quad \frac{dx}{dy} - \left(\frac{1}{y}\right)x = 3y \] This fits the alternative linear structure \(\frac{dx}{dy} + P_1(y)x = Q_1(y)\), where \(x\) is treated as the dependent variable. Thus, it is linear. Hence, Option (C) is the only non-linear option.