BEU previous year question Paper , BEU notes , BEU organizer. BEU pyq solution all semester all branch. MATHS PYQ SOLUTION 2022. Electrical engineering maths 2nd semester.

Q1. (a) Reduce the matrix to normal form and find its rank.7A = [[2, 3, 4, 5], [3, 4, 5, 6], [4, 5, 6, 7], [5, 6, 7, 6]]

(B) For what values of k. the equation x + y + x = 1 2x + y + 4z = k4x + y + 10z – k ^ 2 has a solution?

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2. (a) Verify Cayley-Hamilton theorem for the matrix A = det [[2, 1, 1], [0, 1, 0], [1, 1, 2]]. Hence prove that

A ^ 8 – 5A ^ 7 + 7A ^ 6 – 3A ^ 5 + A ^ 4 – 5A ^ 0.7

8A ^ 2 – 2A + I = [[8, 5, 5], [0, 3, 0], [5, 5, 8]]

b) Find a matrix P which transforms the [1 0 -17 matrix A-1 2 2 2 Hence find A4. 1 3 to diagonal form.

Q3 (a) Find the smallest positive root ofx ^ 4 – x = 10 by using the Newton-Raphson method, correct tothree decimal places.

(b) Evaluate Delta ^ 2 * ((5x + 12)/(x ^ 2 + 5x + 6))differencing being unity.

Q4 (a) Derive the Newton’s forward difference formula using the operator relations.

(b) Using the trapezoidal rule, evaluate the integral I = integrate 1/(x ^ 2 + 6x + 10) dx from 0 to Lwith 2 and 4sub-intervals. Compare with the exact solution. Comment on the magnitudesof the errors obtained.

Q5. Using the Adams-Bashforth predictor-corrector equations, evaluate y (*4) if ysatisfiesd/dx (y) + y/x = 1/(x ^ 2)andmu(1) = 1u(l * 1) = 0.996 y(l * 2) = 0 , y(1 * 3) = 0.972

Q5. Solve the equationu xx =u 1 subject tou(x, 0) = 0 , u(0, t )=0 and u(1, 1) = t, for two timesteps by Crank-Nicolson method.

Q6. (a) Find the inverse Laplace transform of s/((s ^ 2 + a ^ 2) ^ 3) theorem. with the help of convolution

(b) Find the values of the integral oint 0 (e ^ (- 3t) * sin^2 t)/t * c transform. with the help of Laplace

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Q9.Find the solution of following differential equations by using Laplace transform: 7 * 2 = 1

(a) t * (d ^ 2 * y)/(d * t ^ 2) + (1 – 2t) * d/dt (y) – 2y = 0 when

y(0) = 1 y’ * (0) = 2

(b) partial^ 2 y partial t ^ 2 =9 partial^ 2 y partial x ^ 2 , where y(0, t )=0, y2, t )=0 and y(x, 0) = 20sin 2pi*x – 10sin 5pi*x

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