Question:

Which one of the following equations is a linear differential equation?

Show Hint

In a linear differential equation, the dependent variable and its derivatives must appear only to the first power, and there should be no products or nonlinear combinations of them.
Updated On: May 2, 2025
  • \( \frac{dy}{dx} + 2x = y^2 \)
  • \( x^3 \frac{dy}{dx} + xy = x^2 \)
  • \( x^2 \frac{d^2y}{dx^2} + 2y \frac{dy}{dx} = 0 \)
  • \( \left( \frac{dy}{dx} \right)^2 + 2x = y \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

To determine which of the given equations is linear, let's recall the definition of a linear differential equation. A linear differential equation is one in which the dependent variable and its derivatives appear to the first power, and there are no products or nonlinear combinations of the dependent variable or its derivatives. Let’s examine each option:
- Option (A): \( \frac{dy}{dx} + 2x = y^2 \)
This is a non-linear differential equation because the dependent variable \( y \) appears as \( y^2 \), which is a non-linear term. Therefore, option (A) is not linear.
- Option (B): \( x^3 \frac{dy}{dx} + xy = x^2 \)
This is a linear differential equation. Both \( y \) and its derivative \( \frac{dy}{dx} \) appear to the first power, and there are no products or non-linear terms involving \( y \) or its derivatives. Therefore, option (B) is linear.
- Option (C): \( x^2 \frac{d^2y}{dx^2} + 2y \frac{dy}{dx} = 0 \)
This is a non-linear equation because the term \( 2y \frac{dy}{dx} \) involves a product of the dependent variable \( y \) and its derivative \( \frac{dy}{dx} \). Hence, option (C) is non-linear.
- Option (D): \( \left( \frac{dy}{dx} \right)^2 + 2x = y \)
This is a non-linear equation because the derivative \( \frac{dy}{dx} \) appears squared. Hence, option (D) is also non-linear.
Thus, the only linear differential equation is option (B), which makes it the correct answer.
Was this answer helpful?
0
0

Questions Asked in GATE PI exam

View More Questions