Division of an integral into multiple integrals MCQ Quiz - Objective Question with Answer for Division of an integral into multiple integrals - Download Free PDF

Last updated on May 20, 2025

Latest Division of an integral into multiple integrals MCQ Objective Questions

Division of an integral into multiple integrals Question 1:

If [(x + 1)2 f(x) – g(x)] is the particular integral of the differential equation shown below, then

(x+1)2d2ydx23(x+1)dydx+4y=x2

  1. f(x) = log (x + 1)2 and g(x) = 2x + 5
  2. f(x)=log(x+1)22 and g(x)=2x+13
  3. f(x) = log (x + 1) and g(x) = 5x + 1
  4. f(x)=12[log(x+1)]2 and  g(x)=8x+74

Answer (Detailed Solution Below)

Option 4 : f(x)=12[log(x+1)]2 and  g(x)=8x+74

Division of an integral into multiple integrals Question 1 Detailed Solution

Let x + 1 = ez so that z = log (x + 1) and x = ez – 1.

The given equation becomes:

[D(D - 1) – 3D + 4]y = (ez - 1)2, d = d/dz

Or (D2 – 4D + 4)y = e2z – 2ez + 1

The auxiliary equation is m2 – 4m + 4 = 0

(m – 2)2 = 0

∴ C.F.=[(C1+C2z)e2z=[C1+C2log(x+1)](x+1)2

P.I.=1(D2)2e2z2(D2)2ez+1(D2)20.z

z22e22(12)2ez+1(02)2e0.z

z22e2z2ez+14

12(x+1)2{log(x+1)}22(x+1)+14

The correct answer is:

f(x)=12[log(x+1)]2 and

g(x)=8x+74

Top Division of an integral into multiple integrals MCQ Objective Questions

If [(x + 1)2 f(x) – g(x)] is the particular integral of the differential equation shown below, then

(x+1)2d2ydx23(x+1)dydx+4y=x2

  1. f(x) = log (x + 1)2 and g(x) = 2x + 5
  2. f(x)=log(x+1)22 and g(x)=2x+13
  3. f(x) = log (x + 1) and g(x) = 5x + 1
  4. f(x)=12[log(x+1)]2 and  g(x)=8x+74

Answer (Detailed Solution Below)

Option 4 : f(x)=12[log(x+1)]2 and  g(x)=8x+74

Division of an integral into multiple integrals Question 2 Detailed Solution

Download Solution PDF

Let x + 1 = ez so that z = log (x + 1) and x = ez – 1.

The given equation becomes:

[D(D - 1) – 3D + 4]y = (ez - 1)2, d = d/dz

Or (D2 – 4D + 4)y = e2z – 2ez + 1

The auxiliary equation is m2 – 4m + 4 = 0

(m – 2)2 = 0

∴ C.F.=[(C1+C2z)e2z=[C1+C2log(x+1)](x+1)2

P.I.=1(D2)2e2z2(D2)2ez+1(D2)20.z

z22e22(12)2ez+1(02)2e0.z

z22e2z2ez+14

12(x+1)2{log(x+1)}22(x+1)+14

The correct answer is:

f(x)=12[log(x+1)]2 and

g(x)=8x+74

Division of an integral into multiple integrals Question 3:

If [(x + 1)2 f(x) – g(x)] is the particular integral of the differential equation shown below, then

(x+1)2d2ydx23(x+1)dydx+4y=x2

  1. f(x) = log (x + 1)2 and g(x) = 2x + 5
  2. f(x)=log(x+1)22 and g(x)=2x+13
  3. f(x) = log (x + 1) and g(x) = 5x + 1
  4. f(x)=12[log(x+1)]2 and  g(x)=8x+74

Answer (Detailed Solution Below)

Option 4 : f(x)=12[log(x+1)]2 and  g(x)=8x+74

Division of an integral into multiple integrals Question 3 Detailed Solution

Let x + 1 = ez so that z = log (x + 1) and x = ez – 1.

The given equation becomes:

[D(D - 1) – 3D + 4]y = (ez - 1)2, d = d/dz

Or (D2 – 4D + 4)y = e2z – 2ez + 1

The auxiliary equation is m2 – 4m + 4 = 0

(m – 2)2 = 0

∴ C.F.=[(C1+C2z)e2z=[C1+C2log(x+1)](x+1)2

P.I.=1(D2)2e2z2(D2)2ez+1(D2)20.z

z22e22(12)2ez+1(02)2e0.z

z22e2z2ez+14

12(x+1)2{log(x+1)}22(x+1)+14

The correct answer is:

f(x)=12[log(x+1)]2 and

g(x)=8x+74

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