IIT JAM Mathematics Test Series 2023 : Differential Equations – Method of Variation of Parameters

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IIT JAM Mathematics Test Series 2023 : Differential Equations – Method of Variation of Parameters


Q1. Apply the method of variation of parameters to solve

y2 + a2y = cosec ax

  • (a) y = C1 cos ax + C2 sin ax – (x/a) × cos ax + (1/a2) × cos ax log sin ax
  • (b) y = C1 sin ax + C2 cos ax – (x/a) × cos ax + (1/a2) × sin ax log sin ax
  • (c) y = C1 cos ax + C2 sin ax – (x/a) × cos ax + (1/a2) × cos ax log cos ax
  • (d) y = C1 cos ax + C2 sin ax – (x/a) × sin ax + (1/a2) × sin ax log cos ax

Q2. Apply the method of variation of parameters to solve

y2 – 4y1 + 3y = ex/(1 + ex)

  • (a) y = C1 e-x + C2 e3x + (1/2) × (ex– e3x) log (1 – e–x) + (1/2) × e2x
  • (b) y = C1 ex + C2 e3x + (1/2) × (ex– e3x) log (1 + e–x) + (1/2) × e2x
  • (c) y = C1 ex + C2 e3x + (1/2) × (ex– e3x) log (1 – e–x) + (1/2) × e-2x
  • (d) y = C1 e-x + C2 e3x + (1/2) × (ex– e3x) log (1 + e–x) + (1/2) × e2x

Q3. Using method of variation of parameters, solve

d2y/dx2 – 2(dy/dx) + y = x ex sin x

with y (0) = 0 and (dy/dx)x=0 = 0.

  • (a) y = ex (2 – x sin x – 2 cos x).
  • (b) y = ex (2 + x sin x – 2 sin x).
  • (c) y = ex (2 – x sin x – 2 cos x).
  • (d) y = ex (2 + x cos x – 2 cos x).

Q4. Apply the method of variation of parameters to solve

x2y2 + xy1 – y = x2ex

  • (a) y = C1 x + C2 x–1 + ex (1 + x–1)
  • (b) y = C1 – C2 x–1 – ex (1 – x–1)
  • (c) y = C1 x + C2 – ex (1 + x–1)
  • (d) y = C1 x + C2 x–1 + ex (1 – x–1)

Q5. Apply the method of variation of parameters to solve

x2 y” – 2xy’ + 2y = x log x, x > 0

  • (a) y = C1x + C2x2 – (x/2) × (log x)2 – x (1 + log x)
  • (b) y = C1x + C2x2 – (x/2) × (log x) + x (1 + log x)
  • (c) y = C1x – C2x2 + (x/2) × (log x) + x (1 + log x)
  • (d) y = C1x + C2x – (x/2) × (log x)2 – x (1 – log x)

Q6. Solve

y2 – 2y1 + y = x ex log x, x > 0

by the method of variation of parameters.

  • (a) y = C1 e + C2 x ex – (1/6) × x3 ex log x – (5/36) × x3 ex
  • (b) y = C1 ex + C2 x ex + (1/6) × x3 ex log x – (5/36) × x3 ex
  • (c) y = C1 e + C2 x ex + (1/6) × x ex log x + (5/36) × x3 ex
  • (d) y = C1 ex + C2 x ex + (1/6) × x ex log x + (5/36) × x3 ex

Q7. Solve the differential equation

(D2– 2D + 2) y = ex tan x

by method of variation of parameters

  • (a) y = ex (C1 cos x – C2 sin x) + ex cos x log (sin x – tan x)
  • (b) y = ex (C1 cos x – C2 sin x) – ex cos x log (sin x – tan x)
  • (c) y = ex (C1 cos x + C2 sin x) – ex cos x log (sin x + tan x)
  • (d) y = ex (C1 cos x – C2 sin x) – ex cos x log (sin x + tan x)

Q8. Apply the method of variation of parameters to solve

y3 – 6y2 + 11y1 – 6y = e2x

  • (a) y = c1ex – c2e2x + c3e3x+ x e2x
  • (b) y = c1ex + c2ex – c3e3x– x e2x
  • (c) y = c1ex – c2e2x + c3ex– x e2x
  • (d) y = c1ex + c2e2x – c3e3x– x e2x

Q9. Find the particular integral of

(d2y/dx2) – 2(dy/dx) + y = 2x

by the method of variation of parameters

  • (a) 2x + 4
  • (b) 4x + 2
  • (c) 2x – 4
  • (d) 4x – 2

Q10. Apply the method of variation of parameters to solve

d2y/dx2 + y = sec3 x

  • (a) y = c1 sin x – c2 sin x + sin tan x
  • (b) y = c1 cos x + c2 sin x + (1/ 2) sin tan x
  • (c) y = c1 sin x – c2 sin x + (1/ 2) sin tan x
  • (d) y = c1 cos x + c2 sin x – (1/ 2) tan sinx

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Answer Key

01.(a)06.(b)
02.(b)07.(c)
03.(c)08.(d)
04.(d)09.(a)
05.(a)10.(b)

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