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

Last updated on Jul 4, 2025

Latest Variation of Parameters MCQ Objective Questions

Variation of Parameters Question 1:

The Given Differential equation is x2y3xy+4y=ln(x),x>0 with conditions  y(1) = 6 and y'(1) = 13  then the value of y(e) is ?

  1. y(e)=6e217e2
  2. y(e)=6e2+17e2
  3. y(e)=7e216e2
  4. y(e)=7e2+16e2

Answer (Detailed Solution Below)

Option 4 : y(e)=7e2+16e2

Variation of Parameters Question 1 Detailed Solution

Explanation:

x2y3xy+4y=ln(x),x>0

It is Cauchy Euler equation 

let  x=ez   and

replacing   x2y=r(r1),xy=r 

r(r-1) - 3r + 4 = 0 

⇒ r24r+4=0 

⇒ (r2)2=0  

The root is a repeated root r = 2

Hence, the solution to the homogeneous equation is:

⇒ C.F=C1x2+C2x2ln(x)  

⇒ yp=u1(x)y1(x)+u2(x)y2(x),  

where  y1(x)=x2  and  y2(x)=x2ln(x) 

W(y1,y2)=|y1y2y1y2|   

⇒ W(y1,y2)=|x2x2ln(x)2x2xln(x)+x|  

⇒ W(y1,y2)=x2(2xln(x)+x)(x2ln(x))(2x)  

⇒ W(y1,y2)=2x3ln(x)+x32x3ln(x)   

⇒ W(y1,y2)=x3   

u1(x)=ln2(x)xdx,u2(x)=ln(x)xdx  

⇒ u2(x)=ln(x)xdx=ln2(x)2   

The particular solution is:

yp=u1(x)y1(x)+u2(x)y2(x)  

⇒ yp=(ln3(x)3)x2+(ln2(x)2)x2ln(x)  

⇒ yp=x2ln3(x)3+x2ln3(x)2  

⇒ yp=x2ln3(x)(13+12)  

⇒ yp=x2ln3(x)6   

General Solution

y=yh+yp  

Substitute  yh=C1x2+C2x2ln(x)  and  yp=x2ln3(x)6 :

⇒ y=C1x2+C2x2ln(x)+x2ln3(x)6 

and   y=2xC1+C2(2xln(x)+x)+16(2xln3x+3xln2x)

Since y(1) = 6 ⇒  C1=6

Also y'(1) = 13

13=2C1+C2

13=12+C2

C2=1  

Now, General Solution is :

y=6x2+x2ln(x)+x2ln3(x)6  

y(e)=6e2+e2ln(e)+e2×16  

y(e)=7e2+16e2  

Hence Option(4) is the Correct Answer.

Variation of Parameters Question 2:

The Given Differential equation is  y+y=tan(x),0<x<π2 ,with the initial conditions y(0) = 1,  y'(0) = 0 then General Solution is ?

  1. y(x)=cos(x)+ln|sec(x)+tan(x)|cos(x)
  2. y(x)=cos(x)ln|sec(x)+tan(x)|cos(x) 
  3. y(x)=cos(x)+ln|sec(x)+tan(x)|cos(x) 
  4. y(x)=cos(x)ln|sec(x)+tan(x)|cos(x)

Answer (Detailed Solution Below)

Option 4 : y(x)=cos(x)ln|sec(x)+tan(x)|cos(x)

Variation of Parameters Question 2 Detailed Solution

Explanation:

y+y=tan(x),0<x<π2 ,

The characteristic equation is:  

r2+1=0r=±i  

Thus, the general solution to the homogeneous equation is:  

yh=C1cos(x)+C2sin(x),
  
where cos(x) and sin(x) are linearly independent solutions

Particular solution using variation of parameters

Let  yp=u1(x)cos(x)+u2(x)sin(x),
  
where  u1(x)  and  u2(x)  are functions to be determined

W=|cos(x)sin(x)sin(x)cos(x)|=cos2(x)+sin2(x)=1  

Find u1(x) :  

u1(x)=sin(x)tan(x)dx  

u1(x)=sin2(x)cos(x)dx=1cos2(x)cos(x)dx 

u1(x)=1cos(x)dx+cos(x)dx  

u1(x)=ln|sec(x)+tan(x)|+sin(x)  

 

Find u2(x)  :  

u2(x)=cos(x)tan(x)dx

u2(x)=sin(x)dx=cos(x)  

Particular solution:

yp=(ln|sec(x)+tan(x)|+sin(x))cos(x)+(cos(x))sin(x)  

yp=ln|sec(x)+tan(x)|cos(x)+sin(x)cos(x)cos(x)sin(x)  

yp=ln|sec(x)+tan(x)|cos(x) 

The general solution is:  

y(x)=yh+yp=C1cos(x)+C2sin(x)ln|sec(x)+tan(x)|cos(x)  

Apply initial conditions:

Since y(0) = 1 :  

y(0)=C1cos(0)+C2sin(0)ln|sec(0)+tan(0)|cos(0)  

1=C1(1)+C2(0)ln|1+0|(1)  

C1=1  

y'(0) = 0

Differentiate y(x) :  

y(x)=C1sin(x)+C2cos(x)ddx(ln|sec(x)+tan(x)|cos(x))   
  
At x = 0 :  

y(0)=C1sin(0)+C2cos(0)ddx(ln|sec(x)+tan(x)|cos(x))|x=0

0=0+C2(1)0C2=0   

 

Final Solution is :

 y(x)=cos(x)ln|sec(x)+tan(x)|cos(x)

Hence Option(4) is the correct answer.

Variation of Parameters Question 3:

The  Given differential equation is  y2y+y=ett2+1  then y (1)  is  ?

  1. y(t)=C1e+C2ee2ln(2)+eπ4
  2. y(t)=C1eC2ee2ln(2)+eπ4
  3. y(t)=C1eC2ee2ln(2)eπ4
  4. y(t)=C1e+C2ee2ln(2)eπ4

Answer (Detailed Solution Below)

Option 1 : y(t)=C1e+C2ee2ln(2)+eπ4

Variation of Parameters Question 3 Detailed Solution

Explanation:

y2y+y=ett2+1 

The characteristic equation is: 

r22r+1=0(r1)2=0 

in this case: 

yc(t)=C1et+C2tet  

Here, y1(t)=et  and y2(t)=tet  are the independent solutions

The Wronskian of  y1(t)=et and  y2(t)=tet  is:

W(y1,y2)=|y1(t)y2(t)y1(t)y2(t)|  =|ettetetet+tet|  

W=et(et+tet)tet(et)=e2t+te2tte2t=e2t  

⇒  W=e2t 

The particular solution is :

yp(t)=y1(t)y2(t)g(t)Wdt+y2(t)y1(t)g(t)Wdt,  

 

where  g(t)=ett2+1

Substitute y1(t)=et,y2(t)=tet,W=e2t , and g(t) :

yp(t)=ettetett2+1e2tdt+tetetett2+1e2tdt  

yp(t)=ettt2+1dt+tet1t2+1dt  

Since tt2+1dt=12ln(t2+1) 

Also 1t2+1dt=arctan(t) 

yp(t)=et12ln(t2+1)+tetarctan(t)  

Thus:  yp(t)=et2ln(t2+1)+tetarctan(t)

The general solution is the sum of the complementary and particular solutions: 

y(t)=yc(t)+yp(t)  

y(t)=C1et+C2tetet2ln(t2+1)+tetarctan(t)  

 

This is the general solution.

y(t)=C1e+C2ee2ln(2)+eπ4  

Hence Option(1) is the correct answer.

Top Variation of Parameters MCQ Objective Questions

Variation of Parameters Question 4:

The Given Differential equation is  y+y=tan(x),0<x<π2 ,with the initial conditions y(0) = 1,  y'(0) = 0 then General Solution is ?

  1. y(x)=cos(x)+ln|sec(x)+tan(x)|cos(x)
  2. y(x)=cos(x)ln|sec(x)+tan(x)|cos(x) 
  3. y(x)=cos(x)+ln|sec(x)+tan(x)|cos(x) 
  4. y(x)=cos(x)ln|sec(x)+tan(x)|cos(x)

Answer (Detailed Solution Below)

Option 4 : y(x)=cos(x)ln|sec(x)+tan(x)|cos(x)

Variation of Parameters Question 4 Detailed Solution

Explanation:

y+y=tan(x),0<x<π2 ,

The characteristic equation is:  

r2+1=0r=±i  

Thus, the general solution to the homogeneous equation is:  

yh=C1cos(x)+C2sin(x),
  
where cos(x) and sin(x) are linearly independent solutions

Particular solution using variation of parameters

Let  yp=u1(x)cos(x)+u2(x)sin(x),
  
where  u1(x)  and  u2(x)  are functions to be determined

W=|cos(x)sin(x)sin(x)cos(x)|=cos2(x)+sin2(x)=1  

Find u1(x) :  

u1(x)=sin(x)tan(x)dx  

u1(x)=sin2(x)cos(x)dx=1cos2(x)cos(x)dx 

u1(x)=1cos(x)dx+cos(x)dx  

u1(x)=ln|sec(x)+tan(x)|+sin(x)  

 

Find u2(x)  :  

u2(x)=cos(x)tan(x)dx

u2(x)=sin(x)dx=cos(x)  

Particular solution:

yp=(ln|sec(x)+tan(x)|+sin(x))cos(x)+(cos(x))sin(x)  

yp=ln|sec(x)+tan(x)|cos(x)+sin(x)cos(x)cos(x)sin(x)  

yp=ln|sec(x)+tan(x)|cos(x) 

The general solution is:  

y(x)=yh+yp=C1cos(x)+C2sin(x)ln|sec(x)+tan(x)|cos(x)  

Apply initial conditions:

Since y(0) = 1 :  

y(0)=C1cos(0)+C2sin(0)ln|sec(0)+tan(0)|cos(0)  

1=C1(1)+C2(0)ln|1+0|(1)  

C1=1  

y'(0) = 0

Differentiate y(x) :  

y(x)=C1sin(x)+C2cos(x)ddx(ln|sec(x)+tan(x)|cos(x))   
  
At x = 0 :  

y(0)=C1sin(0)+C2cos(0)ddx(ln|sec(x)+tan(x)|cos(x))|x=0

0=0+C2(1)0C2=0   

 

Final Solution is :

 y(x)=cos(x)ln|sec(x)+tan(x)|cos(x)

Hence Option(4) is the correct answer.

Variation of Parameters Question 5:

The Given Differential equation is x2y3xy+4y=ln(x),x>0 with conditions  y(1) = 6 and y'(1) = 13  then the value of y(e) is ?

  1. y(e)=6e217e2
  2. y(e)=6e2+17e2
  3. y(e)=7e216e2
  4. y(e)=7e2+16e2

Answer (Detailed Solution Below)

Option 4 : y(e)=7e2+16e2

Variation of Parameters Question 5 Detailed Solution

Explanation:

x2y3xy+4y=ln(x),x>0

It is Cauchy Euler equation 

let  x=ez   and

replacing   x2y=r(r1),xy=r 

r(r-1) - 3r + 4 = 0 

⇒ r24r+4=0 

⇒ (r2)2=0  

The root is a repeated root r = 2

Hence, the solution to the homogeneous equation is:

⇒ C.F=C1x2+C2x2ln(x)  

⇒ yp=u1(x)y1(x)+u2(x)y2(x),  

where  y1(x)=x2  and  y2(x)=x2ln(x) 

W(y1,y2)=|y1y2y1y2|   

⇒ W(y1,y2)=|x2x2ln(x)2x2xln(x)+x|  

⇒ W(y1,y2)=x2(2xln(x)+x)(x2ln(x))(2x)  

⇒ W(y1,y2)=2x3ln(x)+x32x3ln(x)   

⇒ W(y1,y2)=x3   

u1(x)=ln2(x)xdx,u2(x)=ln(x)xdx  

⇒ u2(x)=ln(x)xdx=ln2(x)2   

The particular solution is:

yp=u1(x)y1(x)+u2(x)y2(x)  

⇒ yp=(ln3(x)3)x2+(ln2(x)2)x2ln(x)  

⇒ yp=x2ln3(x)3+x2ln3(x)2  

⇒ yp=x2ln3(x)(13+12)  

⇒ yp=x2ln3(x)6   

General Solution

y=yh+yp  

Substitute  yh=C1x2+C2x2ln(x)  and  yp=x2ln3(x)6 :

⇒ y=C1x2+C2x2ln(x)+x2ln3(x)6 

and   y=2xC1+C2(2xln(x)+x)+16(2xln3x+3xln2x)

Since y(1) = 6 ⇒  C1=6

Also y'(1) = 13

13=2C1+C2

13=12+C2

C2=1  

Now, General Solution is :

y=6x2+x2ln(x)+x2ln3(x)6  

y(e)=6e2+e2ln(e)+e2×16  

y(e)=7e2+16e2  

Hence Option(4) is the Correct Answer.

Variation of Parameters Question 6:

The  Given differential equation is  y2y+y=ett2+1  then y (1)  is  ?

  1. y(t)=C1e+C2ee2ln(2)+eπ4
  2. y(t)=C1eC2ee2ln(2)+eπ4
  3. y(t)=C1eC2ee2ln(2)eπ4
  4. y(t)=C1e+C2ee2ln(2)eπ4

Answer (Detailed Solution Below)

Option 1 : y(t)=C1e+C2ee2ln(2)+eπ4

Variation of Parameters Question 6 Detailed Solution

Explanation:

y2y+y=ett2+1 

The characteristic equation is: 

r22r+1=0(r1)2=0 

in this case: 

yc(t)=C1et+C2tet  

Here, y1(t)=et  and y2(t)=tet  are the independent solutions

The Wronskian of  y1(t)=et and  y2(t)=tet  is:

W(y1,y2)=|y1(t)y2(t)y1(t)y2(t)|  =|ettetetet+tet|  

W=et(et+tet)tet(et)=e2t+te2tte2t=e2t  

⇒  W=e2t 

The particular solution is :

yp(t)=y1(t)y2(t)g(t)Wdt+y2(t)y1(t)g(t)Wdt,  

 

where  g(t)=ett2+1

Substitute y1(t)=et,y2(t)=tet,W=e2t , and g(t) :

yp(t)=ettetett2+1e2tdt+tetetett2+1e2tdt  

yp(t)=ettt2+1dt+tet1t2+1dt  

Since tt2+1dt=12ln(t2+1) 

Also 1t2+1dt=arctan(t) 

yp(t)=et12ln(t2+1)+tetarctan(t)  

Thus:  yp(t)=et2ln(t2+1)+tetarctan(t)

The general solution is the sum of the complementary and particular solutions: 

y(t)=yc(t)+yp(t)  

y(t)=C1et+C2tetet2ln(t2+1)+tetarctan(t)  

 

This is the general solution.

y(t)=C1e+C2ee2ln(2)+eπ4  

Hence Option(1) is the correct answer.

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