Question:

Which of the following differential equations is/are nonlinear?

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For differential equations: 1. Linear equations involve \( x(t) \) and its derivatives appearing linearly. 2. Nonlinear equations include products, powers, or transcendental functions of \( x(t) \) or its derivatives.
Updated On: Jan 23, 2025
  • \( t x(t) + \frac{dx(t)}{dt} = t^2 e^t, \, x(0) = 0 \)
  • \( \frac{1}{2} e^t + x(t) \frac{dx(t)}{dt} = 0, \, x(0) = 0 \)
  • \( x(t) \cos t - \frac{dx(t)}{dt} \sin t = 1, \, x(0) = 0 \)
  • \( x(t) + e^{\frac{dx(t)}{dt}} = 1, \, x(0) = 0 \)
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The Correct Option is B

Solution and Explanation

Step 1: Analyzing each option. From Option (1): The equation is \( t x(t) + \frac{dx(t)}{dt} = t^2 e^t \). - The equation is linear because \( x(t) \) and its derivative \( \frac{dx(t)}{dt} \) appear linearly (no powers, products, or transcendental functions). From Option (2): The equation is \( \frac{1}{2} e^t + x(t) \frac{dx(t)}{dt} = 0 \). - The term \( x(t) \frac{dx(t)}{dt} \) involves the product of \( x(t) \) and its derivative, making it nonlinear. From Option (3): The equation is \( x(t) \cos t - \frac{dx(t)}{dt} \sin t = 1 \). - The equation is linear because \( x(t) \) and \( \frac{dx(t)}{dt} \) appear linearly with no products or nonlinear operations. From Option (4): The equation is \( x(t) + e^{\frac{dx(t)}{dt}} = 1 \). - The term \( e^{\frac{dx(t)}{dt}} \) involves the exponential of the derivative, making the equation nonlinear. Step 2: Final Answer. The nonlinear equations are (2) \( \frac{1}{2} e^t + x(t) \frac{dx(t)}{dt} = 0 \) and (4) \( x(t) + e^{\frac{dx(t)}{dt}} = 1 \).
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