Method of Variation of Parameters MCQ Quiz - Objective Question with Answer for Method of Variation of Parameters - Download Free PDF

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Latest Method of Variation of Parameters MCQ Objective Questions

Method of Variation of Parameters Question 1:

The differential equation y6y+9y=e3xx2 is solving by the method of variation of parameters, where the complementary function is given by- y = c1y1(x) + c2y2(x)

  1. The value of y1(x) = e3x
  2. The value of ​y2(x) = e3x
  3. The value of ​y2(x) = xe3x
  4. W=|y1y2y1y2|=e6x

Answer (Detailed Solution Below)

Option :

Method of Variation of Parameters Question 1 Detailed Solution

Concept:

Roots of Auxiliary Equation

Complementary Function

m1, m2, m3, … (real and different roots)

C1em1x+C2em2x+C3em3x+

m1, m1, m3, … (two real and equal roots)

(C1+C2x)em1x+C3em3x+

m1, m1, m1, m4… (three real and equal roots)

(C1+C2x+C3x2)em1x+C4em4x+

α + iβ, α – iβ, m3, … (a pair of imaginary roots)

eαx(C1cosβx+C2sinβx)+C3em3x+

α ± iβ, α ± iβ, m5, … (two pairs of equal imaginary roots)

eαx((C1+C2x)cosβx+(C3+C4x)sinβx)+C5em5x+

 

Calculation:

Given:

(D2 – 6D + 9)y = e3x/x2

Auxiliary equation is

(D2 – 6D + 9) = 0

⇒ (D – 3)2 = 0

C.F. = (C1 + C2x) e3x

C.F = C1(e3x) + C2(xe3x)

Wronskian, W=|y1y2y11y21|=|e3xxe3x3e3xe3x+3xe3x|=e6x

Method of Variation of Parameters Question 2:

The Wronskian of the differential equation d2ydx23dydx+2y=ex1+ex is

  1. ex
  2. -e3x
  3. e2x
  4. e5x

Answer (Detailed Solution Below)

Option 2 : -e3x

Method of Variation of Parameters Question 2 Detailed Solution

If the given differential equation is in the form, y'' + py' + qy = x

Where, p, q, x are functions of x.

Then Wronskian is, W=|y1y2y1y2|

Here, y1 and y2 are the solutions of y'' + py' + qy = 0

Given differential equation is,

(D23D+2)y=ex1+ex

Auxiliary equation is,

D2 – 3D + 2 = 0

⇒ D = 1, 2

C.F. = c1 e2x + c2 ex

y1 = e2x, y2 = ex

W=|e2xex2e2xex|=e3x

Method of Variation of Parameters Question 3:

Solution of the differential equation is dydx=sin(x+y)+cos(x+y)

  1. x=ln(1+tan12(x+y))+c
  2. x=ln(tan(x+y))+c
  3. x=ln(x+y+c)
  4. x=tan(x+y+1)+c

Answer (Detailed Solution Below)

Option 1 : x=ln(1+tan12(x+y))+c

Method of Variation of Parameters Question 3 Detailed Solution

Explanation:

x+y=tdydx=dtdx1

dydx=sin(x+y)+cos(x+y)

dtdx1=sint+costdtdx=1+cost+sint

dtdx=2cos2t2+2sint2cost2

dtdx=2cos2(t2)(1+tant2)=2(1+tant2)sec2t2

dx=sec2t22(1+tan(t2))dt

x=ln(1+tant2)+c

x=ln(1+tan12(x+y))+c

Method of Variation of Parameters Question 4:

Consider the following differential equation:

x(ydx+xdy)cosyx=y(xdyydx)sinyx

Which of the following is the solution of the above equation (c is an arbitrary constant)?

  1. xycosyx=c
  2. xysinyx=c
  3. xycosyx=c
  4. xysinyx=c

Answer (Detailed Solution Below)

Option 3 : xycosyx=c

Method of Variation of Parameters Question 4 Detailed Solution

Given differential eqaution is,

x(ydx+xdy)cosyx=y(xydydx)sinyxydx+xdyxdyydx=yxtanyx........(1)

Let, y = v × x

dy = vdx + xdv

By substituting values of Y and dY in equation 1, we get

vxdx+vxdx+x2dvvxdx+x2dvvxdx=vtanvxdv+2vdxxdv=vtanv1+2vxdxdv=vtanv2vxdxdv=vtanv1

Integrating both sides

2 log x = log |sec v| - log v + log c

x2=csecvvx2yx=csecyxx2yx=csecyxxycosyx=c

Method of Variation of Parameters Question 5:

Find the particular solution of the differential equation d2ydx24dydx+4y=e2xx2 

  1. xe2x1xlogx.e2x
  2. xe2x+1xlogxe2x
  3. logx.e2x1x.e2x
  4. e2x[1+logx]

Answer (Detailed Solution Below)

Option 4 : e2x[1+logx]

Method of Variation of Parameters Question 5 Detailed Solution

Given differential equation is (D24D+4)y=e2xx2 

auxiliary equation is (D2)2=0D=2,2

yc=(c1+c2x)e2x=c1e2x+c2xe2x=c1y1+c2y2

This problem can be solved by using the method of variation of parameters .Then yp=Ay1+By2

A=xy2ω.dx,B=xy1ω.dx

ω=|y1y2y11y21|=|e2xxe2x2e2xe2x+2xe2x|=e4x

A=e2xx2.xe2xe4xdx=1x.dx=logx

B=e2xx2×e2xe4x.dx=1x2.dx=1x

yp=logx.e2x1xxe2x=e2x[1+logx]

Top Method of Variation of Parameters MCQ Objective Questions

Consider the following differential equation:

x(ydx+xdy)cosyx=y(xdyydx)sinyx

Which of the following is the solution of the above equation (c is an arbitrary constant)?

  1. xycosyx=c
  2. xysinyx=c
  3. xycosyx=c
  4. xysinyx=c

Answer (Detailed Solution Below)

Option 3 : xycosyx=c

Method of Variation of Parameters Question 6 Detailed Solution

Download Solution PDF

Given differential eqaution is,

x(ydx+xdy)cosyx=y(xydydx)sinyxydx+xdyxdyydx=yxtanyx........(1)

Let, y = v × x

dy = vdx + xdv

By substituting values of Y and dY in equation 1, we get

vxdx+vxdx+x2dvvxdx+x2dvvxdx=vtanvxdv+2vdxxdv=vtanv1+2vxdxdv=vtanv2vxdxdv=vtanv1

Integrating both sides

2 log x = log |sec v| - log v + log c

x2=csecvvx2yx=csecyxx2yx=csecyxxycosyx=c

Method of Variation of Parameters Question 7:

Solution of the differential equation is dydx=sin(x+y)+cos(x+y)

  1. x=ln(1+tan12(x+y))+c
  2. x=ln(tan(x+y))+c
  3. x=ln(x+y+c)
  4. x=tan(x+y+1)+c

Answer (Detailed Solution Below)

Option 1 : x=ln(1+tan12(x+y))+c

Method of Variation of Parameters Question 7 Detailed Solution

Explanation:

x+y=tdydx=dtdx1

dydx=sin(x+y)+cos(x+y)

dtdx1=sint+costdtdx=1+cost+sint

dtdx=2cos2t2+2sint2cost2

dtdx=2cos2(t2)(1+tant2)=2(1+tant2)sec2t2

dx=sec2t22(1+tan(t2))dt

x=ln(1+tant2)+c

x=ln(1+tan12(x+y))+c

Method of Variation of Parameters Question 8:

Find the particular solution of the differential equation d2ydx24dydx+4y=e2xx2 

  1. xe2x1xlogx.e2x
  2. xe2x+1xlogxe2x
  3. logx.e2x1x.e2x
  4. e2x[1+logx]

Answer (Detailed Solution Below)

Option 4 : e2x[1+logx]

Method of Variation of Parameters Question 8 Detailed Solution

Given differential equation is (D24D+4)y=e2xx2 

auxiliary equation is (D2)2=0D=2,2

yc=(c1+c2x)e2x=c1e2x+c2xe2x=c1y1+c2y2

This problem can be solved by using the method of variation of parameters .Then yp=Ay1+By2

A=xy2ω.dx,B=xy1ω.dx

ω=|y1y2y11y21|=|e2xxe2x2e2xe2x+2xe2x|=e4x

A=e2xx2.xe2xe4xdx=1x.dx=logx

B=e2xx2×e2xe4x.dx=1x2.dx=1x

yp=logx.e2x1xxe2x=e2x[1+logx]

Method of Variation of Parameters Question 9:

Consider the following differential equation:

x(ydx+xdy)cosyx=y(xdyydx)sinyx

Which of the following is the solution of the above equation (c is an arbitrary constant)?

  1. xycosyx=c
  2. xysinyx=c
  3. xycosyx=c
  4. xysinyx=c

Answer (Detailed Solution Below)

Option 3 : xycosyx=c

Method of Variation of Parameters Question 9 Detailed Solution

Given differential eqaution is,

x(ydx+xdy)cosyx=y(xydydx)sinyxydx+xdyxdyydx=yxtanyx........(1)

Let, y = v × x

dy = vdx + xdv

By substituting values of Y and dY in equation 1, we get

vxdx+vxdx+x2dvvxdx+x2dvvxdx=vtanvxdv+2vdxxdv=vtanv1+2vxdxdv=vtanv2vxdxdv=vtanv1

Integrating both sides

2 log x = log |sec v| - log v + log c

x2=csecvvx2yx=csecyxx2yx=csecyxxycosyx=c

Method of Variation of Parameters Question 10:

solution of differential equation (D2+4)y=cosec 2x

  1. y=c1cos2x+c2sinx+(logsin2x)2sin2x

  2. y=(c1x2)cos2x+(c2+logsin2x4)sin2x

  3. y=(c1x2)cos2x+(c2+logsin2x2)sin2x

  4. none of these

Answer (Detailed Solution Below)

Option 2 :

y=(c1x2)cos2x+(c2+logsin2x4)sin2x

Method of Variation of Parameters Question 10 Detailed Solution

Auxiliary equation:

m2+4=0

m=±2i

then CF=c1 cos2x     u(x)   c1 sin2x        v(x)

then dudx=2sin2x, dvdx=2cos2x

By wrongskion,

udvdxvdudx=cos2x(+2cos2x)(sin2x)(2sin2x)

=2(cos22x+sin22x)=2

PI = A u(x) + B v(x)

where A=v(x)R(x)vdudxvdudz=sin2xcosec2x2dx=x2

B=u(x)R(x)ududxvdudxdx=+cos2xcosec2x2dx=cot2x2dx=12log(sin2x)2=14log(sin2x)

then the solution is y=CF+PI=yc+yp

y=c1cos2x+c2sin2xx2cos2x+log(sin2x)4sin2x=(c1x2)cos2x+(c2+logsin2x4)sin2x

Method of Variation of Parameters Question 11:

The differential equation d2ydx2y=21+ex is solving by the method of variation of parameters, the Wronskian will be______

  1. -2
  2. -4
  3. 4
  4. -8

Answer (Detailed Solution Below)

Option 1 : -2

Method of Variation of Parameters Question 11 Detailed Solution

Concept:

If the given differential equation is in the form, y'' + py' + qy = x where, p, q, x are functions of x.

Then Wronskian is, W=|y1y2y1y2| where, y1 and y2 are the solutions of y'' + py' + qy = 0

Explanation:

d2ydx2y=2(1+ex)(D21)y=21+ex

Auxiliary equation, D2 – 1 = 0

⇒ D = ± 1

C.F.=C1ex+C2ex

Here, y1 = ex

y2 = e-x

y1=exy2=ex

Wronskian, W=|y1y2y1y2|

=|exexexex|=ex.(ex)ex.ex

= -1-1 = -2

Method of Variation of Parameters Question 12:

The differential equation y6y+9y=e3xx2 is solving by the method of variation of parameters, then Wronskian will be –

Wronskian for solution y = c1y1(t) + c2y2(t) is defined as W(y1,y2)(x)=|y1y2y1y2|

  1. e3x
  2. e6x
  3. e9x
  4. e-3x

Answer (Detailed Solution Below)

Option 2 : e6x

Method of Variation of Parameters Question 12 Detailed Solution

Give equation is (D2 – 6D + 9)y = e3x/x2

Auxiliary equation is, (D2 – 6D + 9) = 0

⇒ (D – 3)2 = 0

C.F. = (C1 + C2x) e3x

Wronskian, W=|y1y2y11y21|=|e3xxe3x3e3xe3x+3xe3x|=e6x

Method of Variation of Parameters Question 13:

The Wronskian of the differential equation d2ydx23dydx+2y=ex1+ex is

  1. ex
  2. -e3x
  3. e2x
  4. e5x

Answer (Detailed Solution Below)

Option 2 : -e3x

Method of Variation of Parameters Question 13 Detailed Solution

If the given differential equation is in the form, y'' + py' + qy = x

Where, p, q, x are functions of x.

Then Wronskian is, W=|y1y2y1y2|

Here, y1 and y2 are the solutions of y'' + py' + qy = 0

Given differential equation is,

(D23D+2)y=ex1+ex

Auxiliary equation is,

D2 – 3D + 2 = 0

⇒ D = 1, 2

C.F. = c1 e2x + c2 ex

y1 = e2x, y2 = ex

W=|e2xex2e2xex|=e3x

Method of Variation of Parameters Question 14:

The differential equation y6y+9y=e3xx2 is solving by the method of variation of parameters, where the complementary function is given by- y = c1y1(x) + c2y2(x)

  1. The value of y1(x) = e3x
  2. The value of ​y2(x) = e3x
  3. The value of ​y2(x) = xe3x
  4. W=|y1y2y1y2|=e6x

Answer (Detailed Solution Below)

Option :

Method of Variation of Parameters Question 14 Detailed Solution

Concept:

Roots of Auxiliary Equation

Complementary Function

m1, m2, m3, … (real and different roots)

C1em1x+C2em2x+C3em3x+

m1, m1, m3, … (two real and equal roots)

(C1+C2x)em1x+C3em3x+

m1, m1, m1, m4… (three real and equal roots)

(C1+C2x+C3x2)em1x+C4em4x+

α + iβ, α – iβ, m3, … (a pair of imaginary roots)

eαx(C1cosβx+C2sinβx)+C3em3x+

α ± iβ, α ± iβ, m5, … (two pairs of equal imaginary roots)

eαx((C1+C2x)cosβx+(C3+C4x)sinβx)+C5em5x+

 

Calculation:

Given:

(D2 – 6D + 9)y = e3x/x2

Auxiliary equation is

(D2 – 6D + 9) = 0

⇒ (D – 3)2 = 0

C.F. = (C1 + C2x) e3x

C.F = C1(e3x) + C2(xe3x)

Wronskian, W=|y1y2y11y21|=|e3xxe3x3e3xe3x+3xe3x|=e6x

Method of Variation of Parameters Question 15:

Solve (x2 – y2) dx – xydy = 0

  1. x2(x2+2y2)=constant
  2. x2(x2+3y2)=constant
  3. x2(x22y2)=constant
  4. x2(2x22y2)=constant

Answer (Detailed Solution Below)

Option 3 : x2(x22y2)=constant

Method of Variation of Parameters Question 15 Detailed Solution

(x2 – y2) dx – xydy = 0

dydx=x2y2xy

This homogeneous in x & y

Put y = vx then

dydx=v+xdvdx

V+xdvdx=1v2v

xdvdx=12v2v

Separating the variables

v12v2dv=dxx

Integrating on both side

v12y2dv=dxx+c

14log(12v2)=logx+c

4logx+log(12v2)=4c

x4(12y2x2)=e3c

⇒ x2 (x2 – 2y2) = constant

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