Question

Which of the following 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 check linearity, look for terms like \(y^2, (y')^2, y \cdot y', \sin(y), \log(y)\). If any of these exist, the equation is non-linear.
Always check if an equation is linear in \(x\) (where \(x\) is the dependent variable) if it fails the linearity test for \(y\).

Answer 1

The Correct Option is C

Concept:
• A differential equation is linear if the dependent variable (\(y\)) and its derivatives (e.g., \(\frac{dy}{dx}\)) appear only in the first degree and are not multiplied together.
• General linear form (in \(y\)): \(\frac{dy}{dx} + P(x)y = Q(x)\).
• General linear form (in \(x\)): \(\frac{dx}{dy} + P(y)x = Q(y)\).

Step 1: Analyze Option (A)
\((1 + x^2) dy + 2xy dx = \cot x dx\)
Divide by \(dx\): \((1 + x^2) \frac{dy}{dx} + 2xy = \cot x\)
Divide by \((1+x^2)\): \(\frac{dy}{dx} + \left(\frac{2x}{1+x^2}\right)y = \frac{\cot x}{1+x^2}\)
This is in the form \(\frac{dy}{dx} + P(x)y = Q(x)\). It is linear.

Step 2: Analyze Option (B)
\(y + \frac{d}{dx}(xy) = x (\sin x + \log x)\)
Expand using product rule: \(y + (x \frac{dy}{dx} + y) = x (\sin x + \log x)\)
\(x \frac{dy}{dx} + 2y = x (\sin x + \log x)\)
Divide by \(x\): \(\frac{dy}{dx} + \frac{2}{x}y = (\sin x + \log x)\)
This is in the form \(\frac{dy}{dx} + P(x)y = Q(x)\). It is linear.

Step 3: Analyze Option (C)
\(x (1 + y^2) dx - y (1 + x^2) dy = 0\)
Divide by \(dx\) and rearrange:
\(y(1 + x^2) \frac{dy}{dx} = x(1 + y^2)\)
\(\frac{dy}{dx} = \frac{x(1 + y^2)}{y(1 + x^2)}\)
Here, the dependent variable \(y\) appears as \(y^2\). A linear equation must have \(y\) only in the first degree.
Thus, this equation is not linear.

Step 4: Analyze Option (D)
\(y dx - (x + 3y^2) dy = 0\)
Rearrange: \(y \frac{dx}{dy} - x = 3y^2\)
Divide by \(y\): \(\frac{dx}{dy} - \frac{1}{y}x = 3y\)
This is in the form \(\frac{dx}{dy} + P(y)x = Q(y)\), which is linear in \(x\).