Previous Year Questions- Differential Equations | Engineering Mathematics for Electrical Engineering - Electrical Engineering (EE) PDF Download

Q1: Which of the following differential equations is/are nonlinear?      (2024)
(a)
(b)
(c)
(d)
Ans: (b, d)

Q2: Consider the functionf(t) = (max(0, t))² for −∞
ltt
lt ∞, where $\max (𝑎,𝑏)$max(a, b) denotes the maximum of a and b. Which of the following statements is/are true?       (2024)
(a) f(t) is not differentiable.
(b) f(t) is differentiable and its derivative is continuous.
(c) f(t) is differentiable but its derivative is not continuous.
(d) f(t) and its derivative are differentiable.
Ans: (b)
Sol: ∴ f(x) is differential and f′(x) is continuous but f′(x) is not continuous.

Q3: A quadratic function of two variables is given as

The magnitude of the maximum rate of change of the function at the point (1, 1) is ____(Round off to the nearest integer).      (2023)
(a) 10
(b) 12
(c) 8
(d) 16
Ans: (a)
Sol: Given :

Q4: Three points in the x−y plane are (−1, 0.8) (0, 2.2) and (1, 2.8) . The value of the slope of the best fit straight line in the least square sense is ___ (Round off to 2 decimal places).       (2023)
(a) 0.25
(b) 0.5
(c) 0.75
(d) 1
Ans: (d)
Sol: Straight line equation, y = ax + b [Let]
where, a = slope
By lest approximation,
From eqn. (1), we get
⇒ a = 1

Q5: In the following differential equation, the numerically obtained value of y(t), at t = 1, is ___ (Round off to 2 decimal places). 
(2023) 
(a) 0.25
(b) 0.5
(c) 0.75
(d) 0.85
Ans: (b)
Sol: Given :
We know,
Taylor's series

Q6: Consider the initial value problem below. The value of y at x = ln 2. (rounded off to 3 decimal places) is ________ . 
(dy/dx) = 2x - y, y(0) = 1      (2020)
(a) 1.386
(b) 0.886
(c) 0.452
(d) 0.642
Ans: (b)
Sol: Solution,

Q7: The partial differential equation is known as      (2019)
(a) heat equation
(b) wave equation
(c) Poisson's equation
(d) Laplace equation
Ans: (b)

Q8: Consider a system governed by the following equations
The initial conditions are such that Which one of the following is true?       (2018)
(a)
(b)
(c)
(d)
Ans: (c)
Sol: 
Q9: Consider the differential equation  There exists a unique solution for this differential equation when t belongs to the interval      (SET-1 (2017))
(a) (-2, 2)
(b) (-10, 10)
(c) (-10, 2)
(d) (0, 10)
Ans: (a)
Sol: The differentail equation,
Only option (A) not cover 9 and -9. Hence it is correct.

Q10: Let y(x) be the solution of the differential equation  with initial conditions y(0) = 0 and  Then the value of y(1) is _________.        (SET-2 (2016))
(a) 2.50
(b) 5.65
(c) 7.38
(d) 9.36
Ans: (c)
Sol: 
Q11: The solution of the differential equation, for  with initial conditions y(0) = 0 and y'(0) = 1, is (u(t) denotes the unit step function),      (SET-2 (2016))
(a)
(b)
(c)
(d)
Ans: (a)
Sol: The differentail equation is

Q12: A solution of the ordinary differential equation is such that y(0) = 2 and y (1) = The value of (dy/dt) (0) is _______      (SET-1 (2015))
(a) -1
(b) -8
(c) 3
(d) -3
Ans: (d)
Sol: 
Q13: Consider the differential equation  Which of the following is a solution to this differential equation for x > 0 ?       (SET-2 (2014))
(a) e^x
(b) x²
(c) 1/x
(d) In x
Ans: (c)
Sol: One independent solution is (1/x)
Another independent solution is x.

Q14: The solution for the differential equation with initial conditions x(0) = 1 and       (SET-1 (2014))
(a) t² + t + 1
(b) sin3t + (1/3)cos3t + (2/3)
(c) (1/3) sin3t + cos3t 
(d) cos3t  + t
Ans: (c)
Sol: Auxiliary equation is $𝑚2+9=0$m²+9 = 0

Q15: With initial condition x(1) = 0.5, the solution of the differential equation  is      (2012)
(a) x = t - (1/2)
(b) x = t² - (1/2)
(c) x = (t²/2)
(d) x = t/2
Ans: (d)
Sol: The given differential equation is  with initial condition x(t) = 1/2 which is same as  
Which is a linear differential equation So, x = (t/2) is the solution.

Q16: With K as a constant, the possible solution for the first order differential equation dy/dx = e^-3x is      (2011)
(a) -(1/3)e^-3x + K
(b) -(1/3)e^3x + K
(c) -(1/3)e^-3x + K
(d) -3e^-x + K
Ans: (a)
Sol: 
Q17: For the differential equation with initial conditions x(0) = 1 and the solution is      (2010)
(a) x(t) = 2e^-6t - e^-2t
(b) x(t) = 2e^-2t - e^-4t
(c) x(t) = -e^-6t + 2e^-4t
(d) x(t) = e^-2t - e^-4t
Ans: (b)
Sol: 
Q18: A differential equation (dx/dt) = e^2t u(t), has to be solved using trapezoidal rule of integration with a step size h = 0.01s. Function u(t) indicates a unit step function. If x(0-) = 0, then value of x at t = 0.01s will be given by       (2008)
(a) 0.00099
(b) 0.00495
(c) 0.0099
(d) 0.0198
Ans: (c)
Sol: 
Q19: For the equation   the solution x(t) approaches which of the following values as t → ∞?      (2005)
(a) 0
(b) 5/2
(c) 5
(d) 10
Ans: (b)
Sol: 
Q20: The solution of the first order differential equation  (2005)
(a) x(t) = x₀e^-3t
(b) x(t) = x₀e^-3
(c) x(t) = x₀e^-t/3
(d) x(t) = x₀e^-t
Ans: (a)
Sol: