Variable Separable Method MCQ Quiz - Objective Question with Answer for Variable Separable Method - Download Free PDF

Last updated on Apr 4, 2025

Latest Variable Separable Method MCQ Objective Questions

Variable Separable Method Question 1:

The solution of differential equation (x+y)2dydx=a2 is _______.

  1. (y + x) = a tan(yca)
  2. y - x = a tan(y - c)
  3. (y - x) = tan(yca)
  4. a(y - x) = tan(yca)

Answer (Detailed Solution Below)

Option 1 : (y + x) = a tan(yca)

Variable Separable Method Question 1 Detailed Solution

Concept:

Variable Seperable Method:

Consider the first-order differential equation, 

P(y)dydx = Q(x), where Q(x) and P(y) are functions involving x and y only, respectively.

We can solve this by separating variables:

P(y)dydx = Q(x) ⇒ P(y)dy=Q(x)dx

Calculation:

Given, (x+y)2dydx=a2

Let, x + y = t

⇒ 1 + dydx = dtdx

⇒ dydx = dtdx - 1

∴ The differential equation becomes, t2(dtdx1)=a2

⇒ dtdx1=a2t2

⇒ dtdx=a2t2+1=a2+t2t2

⇒ t2a2+t2dt=dx

⇒ a2+t2a2a2+t2dt=dx

⇒ (1a2a2+t2)dt=dx

Integrating on both sides, we get:

(1a2a2+t2)dt=dx

⇒ 1dta21a2+t2dt=dx

⇒ ta2×1atan1(ta)=x+c

⇒ x + y - a tan-1(x+ya) = x + c

⇒ y - a tan-1(x+ya) = c

⇒ tan-1(x+ya) = yca

⇒ x+ya = tan(yca)

⇒ y + x = a tan(yca), where c is the constant of integration. 

∴ The solution of the given differential equation is (y + x) = a tan(yca).

The correct answer is Option 1.

Variable Separable Method Question 2:

Differential equation y3dydx+x3=0, y(0) = 1 has a solution given by y. The value of y(-1) is:

  1. -1
  2. 0
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 2 : 0

Variable Separable Method Question 2 Detailed Solution

Concept:

Variable separable method

∫f(y)dy = ∫f(x)dx

Calculation:

Given:

y3dydx+x3=0

y3dy = -x3dx

Integrating both sides

y44=x44+C

At y(0) = 1

C=14

y44=x44+14

y4 = -x4 + 1

At y(-1)

y4 = -(-1)4 + 1 = 0

Variable Separable Method Question 3:

The solution of x(dydx)=y+xsec(yx) is _____

  1. sin (x/y) = log (cx)
  2. sin (y/x) = log (c/x)
  3. sin (y/x) = log (cx)
  4. sin (x/y) = log (c/x)

Answer (Detailed Solution Below)

Option 3 : sin (y/x) = log (cx)

Variable Separable Method Question 3 Detailed Solution

Explanation:

The given equation is Homogenous,

Put,

y = Vx

dydx=V+xdVdx

Substituting, We get

V+xdvdx=V+secV

xdvdx=secV

cosVdv=dxx

Integrating, we have

sin V = log x + log C

sin(yx)=log(cx)

Variable Separable Method Question 4:

Find the general solution of dydx=[(ex1)sec2yextany]1 

  1. ex = c tan y + 1
  2. sin y = c (1 - ex)
  3. ex cos y = 1
  4. cot y = 1 + ex

Answer (Detailed Solution Below)

Option 1 : ex = c tan y + 1

Variable Separable Method Question 4 Detailed Solution

Concept:

Variable separable differential equation:

f(x) dx + f(y) dy = 0

∫ f(x) dx + ∫ f(y) dy = c

Calculation:

dydx=extany(ex1)sec2y

exex1dxsec2ytanydy=0

exex1dxsec2ytanydy=c

⇒ ln (ex - 1) - ln (tan y) = c

ln(ex1tany)=cex1tany=ec=c

ex = c tan y + 1

Variable Separable Method Question 5:

The general solution of the differential equation 9yy’ + 4x = 0 is, (y=dydx,C=constant)

  1. 9x2 + 4y2 = C

  2. x29y24=C
  3. 4x2 + 9y2 = C
  4. x24y29=C

Answer (Detailed Solution Below)

Option 3 : 4x2 + 9y2 = C

Variable Separable Method Question 5 Detailed Solution

Concept:

The general form of a differential equation is,

dydx=f(x,y) or Mdx + Ndy = 0

Where M and N are functions of x and y.

By variable separable form,

In an equation, if it is possible to collect all terms of x and dx on one side and all the terms of y and dy on the other side, then the variables are said to be separable.

After separating, integrate on both the side of the equation,

f(y)dy=ϕ(x)dx+C

Calculation:

Given,

9yy’ + 4x = 0

Where,

(y=dydx,C=constant)

9ydydx+4x=0

9ydy=4xdx

Integrating on both the side,

9ydy=(4x)dx+C1

9y22=4x22+C1

9y2+4x2=2C1

4x2+9y2=C

Hence the general solution of the given differential equation is 4x2+9y2=C

Top Variable Separable Method MCQ Objective Questions

Which one of the following options contains two solutions of the differential equation dydx=(y1)x?

  1. In |y - 1| = 0.5 x2 + C and y = 1
  2. In {y - 1} = 2x2 + C and y = 1
  3. In |y - 1| = 0.5 x2 + C and y = -1
  4. In |y - 1| = 2x2 + C and y = -1

Answer (Detailed Solution Below)

Option 1 : In |y - 1| = 0.5 x2 + C and y = 1

Variable Separable Method Question 6 Detailed Solution

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dydx=(y1)x     ---(1)

The given equation can be solved using the variable separable method as:

dyy1=xdx

Integrating both sides, we get:

dyy1=xdx

ln(y1)=x22+C     ---(2)

Now, from equation (1), we get:

dydx=0 for y = 1

y = constant = 1 is also a solution to the given differential equation.

An ordinary differential equation is given below:

(dydx)(xlnx)=y

The solution for the above equation is (Note: K denotes a constant in the options)

  1. y = Klnx
  2. y = Kxlnx
  3. y = Kxex
  4. y = Kxe-x

Answer (Detailed Solution Below)

Option 1 : y = Klnx

Variable Separable Method Question 7 Detailed Solution

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Given D.E

(dydx)(xlnx)=y

dyy=dxxlnxvariableseparableDifferentialEquation

Integrating on both sides; then

1ydy=1/xlnxdx

lny=ln[lnx]+lnk;[f1(x)f(x)dx=lnf(x)]

⇒ lnylnK=ln[lnx]

ln[yk]=ln[lnx]yk=lnx

⇒ y=klnx

 

Consider the following differential equation:

 dydt=5y; Initial condition: y = 2 at t = 0

The value of y at t = 3 is

  1. – 5e-10
  2. 2e-10
  3. 2e-15
  4. -15e2

Answer (Detailed Solution Below)

Option 3 : 2e-15

Variable Separable Method Question 8 Detailed Solution

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Explanation:

dydt=5y

dyy=5dt (variables separable form)

Integrating both side we get,

lny=5t+c         _______________(1)

initial condition: y = 2 at t = 0,

from equation (1), 

ln2=5×0+c

c=ln2

lny=5t+ln2

ln(y2)=5t

y=2e5t

now at t = 3, y=2e15

Differential equation y3dydx+x3=0, y(0) = 1 has a solution given by y. The value of y(-1) is:

  1. -1
  2. 0
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 2 : 0

Variable Separable Method Question 9 Detailed Solution

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Concept:

Variable separable method

∫f(y)dy = ∫f(x)dx

Calculation:

Given:

y3dydx+x3=0

y3dy = -x3dx

Integrating both sides

y44=x44+C

At y(0) = 1

C=14

y44=x44+14

y4 = -x4 + 1

At y(-1)

y4 = -(-1)4 + 1 = 0

If y is the solution of the differential equation y3dydx+x3=0,y(0)=1, the value of y (-1) is

  1. -2
  2. -1
  3. 0
  4. 1

Answer (Detailed Solution Below)

Option 3 : 0

Variable Separable Method Question 10 Detailed Solution

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Concept:

y3dydx+x3=0

Given y (0) = 1,      y(-1) = ?

y3dydx+x3=0

y3 dy = -x3dx

y3dy=x3dx+c

y44=x44+C

x44+y44=C      ----(1)

Now given y (0) = 1

C=(14)+(0)24

C=14

y44+x44=14                       {By equation (1)}

Y (-1) =?

y44+(1)44=14

y44+14=14

y44=0y=0

y (-1) = 0

The general solution of the differential equation 9yy’ + 4x = 0 is, (y=dydx,C=constant)

  1. 9x2 + 4y2 = C

  2. x29y24=C
  3. 4x2 + 9y2 = C
  4. x24y29=C

Answer (Detailed Solution Below)

Option 3 : 4x2 + 9y2 = C

Variable Separable Method Question 11 Detailed Solution

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Concept:

The general form of a differential equation is,

dydx=f(x,y) or Mdx + Ndy = 0

Where M and N are functions of x and y.

By variable separable form,

In an equation, if it is possible to collect all terms of x and dx on one side and all the terms of y and dy on the other side, then the variables are said to be separable.

After separating, integrate on both the side of the equation,

f(y)dy=ϕ(x)dx+C

Calculation:

Given,

9yy’ + 4x = 0

Where,

(y=dydx,C=constant)

9ydydx+4x=0

9ydy=4xdx

Integrating on both the side,

9ydy=(4x)dx+C1

9y22=4x22+C1

9y2+4x2=2C1

4x2+9y2=C

Hence the general solution of the given differential equation is 4x2+9y2=C

The solution of the differential equation dydx+y2=0 is

  1. y=1x+c
  2. y=x33+c
  3. y = cex
  4. unsolvable as the equation is non-linear

Answer (Detailed Solution Below)

Option 1 : y=1x+c

Variable Separable Method Question 12 Detailed Solution

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Concept:

Variable separable:

Put all 'x' and 'y' terms separately on both sides of the equation.

Calculation:

Given:

dydx+y2=0

dyy2=dx

dyy2=dx

1y=x+C

y=1x+c

The solution of the differential equation dydxxy at x = 1 and y=3 is 

  1. x – y2 = -2
  2. x + y2 = -4
  3. x2 – y2 = 2
  4. x2 + y2 = 4

Answer (Detailed Solution Below)

Option 4 : x2 + y2 = 4

Variable Separable Method Question 13 Detailed Solution

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Concept:

Variable separable method:

Separate the term of 'x' and 'y' in both sides of the equation.

Calculation:

Given:

dydxxy

ydy=xdx

y22=x22+c

At x = 1 and y=3

32=12+c

c = 2

y22=x22+2

y2+x2=4

Variable Separable Method Question 14:

Which one of the following options contains two solutions of the differential equation dydx=(y1)x?

  1. In |y - 1| = 0.5 x2 + C and y = 1
  2. In {y - 1} = 2x2 + C and y = 1
  3. In |y - 1| = 0.5 x2 + C and y = -1
  4. In |y - 1| = 2x2 + C and y = -1

Answer (Detailed Solution Below)

Option 1 : In |y - 1| = 0.5 x2 + C and y = 1

Variable Separable Method Question 14 Detailed Solution

dydx=(y1)x     ---(1)

The given equation can be solved using the variable separable method as:

dyy1=xdx

Integrating both sides, we get:

dyy1=xdx

ln(y1)=x22+C     ---(2)

Now, from equation (1), we get:

dydx=0 for y = 1

y = constant = 1 is also a solution to the given differential equation.

Variable Separable Method Question 15:

Let y(x) be the solution of the differential equation ddx(xdydx)=x;y(1)=0,dydx|x=1=0 then y(2) is

  1. 34+12ln2
  2. 3412ln2
  3. 34+ln2
  4. 34ln2

Answer (Detailed Solution Below)

Option 2 : 3412ln2

Variable Separable Method Question 15 Detailed Solution

Explanation:

ddx(xdydx)=x

xdydx=x22+c1

dydx|x=1=0

0=12+c1

c1=12

dydx=x212x

y=x2412lnx+c2

y(1) = 0

0=140+c2

c2=14

y=x2412lnx14

y(2)=3412ln2
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