Linear MCQ Quiz - Objective Question with Answer for Linear - Download Free PDF

Last updated on Apr 23, 2025

Latest Linear MCQ Objective Questions

Linear Question 1:

The integrating factor of equation sec2ydydx+xtany=x3 is

  1. ex22
  2. ex22
  3. ex2
  4. ex2

Answer (Detailed Solution Below)

Option 1 : ex22

Linear Question 1 Detailed Solution

Concept:

The standard form of a first-order linear differential equation is,

Form 1: dydx+Py=Q

Where P and Q are the functions of x.

Integrating factor, IF=ePdx

Now, the solution for the above differential equation is,

y(IF)=IF.Qdx+C

Form 2: dxdy+Px=Q

Where P and Q are the functions of y.

Integrating factor, IF=ePdy

Now, the solution for the above differential equation is,

x(IF)=IF.Qdy+C

Calculation:

Given the differential equation is

sec2ydydx+xtany=x3

Put tan y = t

By differentiating with respect to t,

⇒ sec2 y dy = dt

Now, the given equation becomes

dtdx+xt=x3

Now, it is in the form of a first order linear differential equation.

Integrating factor =exdx=ex22

Linear Question 2:

The integrating factor of the differential equation (x21)xdydx+2(2x21)y=5x3 is

  1. x2(x2 - 1)
  2. x2(x2 + 1)
  3. x(x2 - 1)
  4. x(x2 + 1)

Answer (Detailed Solution Below)

Option 1 : x2(x2 - 1)

Linear Question 2 Detailed Solution

Concept:

 dydx+P(x)y=Q(x)

Equation of this type is known as Linear First Order Equation, whose solution is given by: y(I.F.)=(Q×I.F.)dx+c whereI.F.=ePdx

Calculation:

Given:

(x21)xdydx+2(2x21)y=5x3

dydx+2(2x21)x(x21)y=5x3x(x21)--------(1)

By comparing equation (1) with dydx+P(x)y=Q(x)  we get

P=2(2x21)x(x21)

P=4x22x(x+1)(x1)

Solving through partial fraction method

Ax+B(x+1)+C(x1)=4x22x(x+1)(x1)

(x + 1)(x - 1)A + x(x-1)B + x(x + 1)C = 4x2 – 2

Ax2 – A + Bx2 – Bx + Cx2 + Cx = 4x2 – 2

(A + B + C)x2 + (C - B)x – A =  4x2 – 2

Comparing co-efficient of x2, x, and constants we get

A + B + C = 4 ……. (ii)

C – B = 0 

⇒ B = C

A = 2

∵ A + B + C = 4

⇒ 2 + B + C = 4

⇒ B + C = 2

∴ B = C = 1

Ax+B(x+1)+C(x1)=4x22x(x+1)(x1)=P

P=2x+1(x+1)+1(x1)

Pdx=(2x+1(x+1)+1(x1))dx

2dxx+dx(x+1)+dx(x1)

⇒ 2ln x + ln (x + 1) + ln (x - 1)

⇒ ln x2 + ln (x2 - 1)

⇒ ln x2(x2 - 1)

I.F.=ePdx

I.F =elnx2(x21)

⇒ I.F = x2 (x2 - 1)

Linear Question 3:

Consider the differential equation dydx+10y=f(x),x>0, where f(x) is continuous function such that limxf(x)=1. Then the value of limxy(x) is ________.

Answer (Detailed Solution Below) 0.1

Linear Question 3 Detailed Solution

Concept:

The standard form of a first-order linear differential equation is,

Form 1: dydx+Py=Q

Where P and Q are the functions of x.

Integrating factor, IF=ePdx

Now, the solution for the above differential equation is,

y(IF)=IF.Qdx+C

Form 2: dxdy+Px=Q

Where P and Q are the functions of y.

Integrating factor, IF=ePdy

Now, the solution for the above differential equation is,

x(IF)=IF.Qdy+C

Calculation:

dydx+10y=f(x),x>0

It is the first order linear differential equation.

Integrating factor, I.F.=e10dx=e10x

The solution of the differential equation is,

y(x)e10x=e10xf(x)dx+C

y(x)=e10xe10xf(x)dx+Ce10x

limxy(x)=limxe10xf(x)dxe10x+limxCe10x

=limxe10xf(x)dxe10x

By applying the L-Hospitals rule,

=limxe10xf(x)10e10x

=110limxf(x)=110=0.1

Linear Question 4:

The differential equation y=xdydx+x(dy/dx) is

  1. First-order and first-degree
  2. First-order and second-degree
  3. Second-order and first-degree
  4. Second-order and second-degree

Answer (Detailed Solution Below)

Option 2 : First-order and second-degree

Linear Question 4 Detailed Solution

Concept:
Differential equation:

It is an equation that involves differential coefficients or differentials.

Order of differential equation:

It is the order of the highest derivative appearing in that equation.

Degree of the differential  equation:

It is the degree (power) of the highest derivative occurring in that equation after the equation has been expressed in a form, free from radicals and fractional powers as far as the derivatives are concerned.

Explanation:

Given D.E is

y=xdydx+x(dy/dx)

ydydx=x(dydx)2+x

Hence the highest order in the equation is 1 of dydxand the highest degree (power) of the equation is 2 of dydx

Hence the equation is First-order and Second-degree*

Linear Question 5:

Second Derivative of sin(x) with respect to x is

  1. sin (x)
  2. cos (x)
  3. –sin (x)
  4. –cos (x)

Answer (Detailed Solution Below)

Option 3 : –sin (x)

Linear Question 5 Detailed Solution

Concept:

  • ddx(xn)=nxn1

where ‘x’ is an algebraic function

  • ddx(sinx)=cosx
  • ddx(cosx)=sinx
  • The second derivative is represented as
  • d2ydx2=ddx(dydx)

Calculation:

Given:

y = sin x

dydx=ddx(sinx)=cosx

Now second derivative of sin(x) with respect to x is

d2ydx2=ddx(cosx)

d2ydx2=sinx

Top Linear MCQ Objective Questions

The integrating factor of the differential equation (x21)xdydx+2(2x21)y=5x3 is

  1. x2(x2 - 1)
  2. x2(x2 + 1)
  3. x(x2 - 1)
  4. x(x2 + 1)

Answer (Detailed Solution Below)

Option 1 : x2(x2 - 1)

Linear Question 6 Detailed Solution

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Concept:

 dydx+P(x)y=Q(x)

Equation of this type is known as Linear First Order Equation, whose solution is given by: y(I.F.)=(Q×I.F.)dx+c whereI.F.=ePdx

Calculation:

Given:

(x21)xdydx+2(2x21)y=5x3

dydx+2(2x21)x(x21)y=5x3x(x21)--------(1)

By comparing equation (1) with dydx+P(x)y=Q(x)  we get

P=2(2x21)x(x21)

P=4x22x(x+1)(x1)

Solving through partial fraction method

Ax+B(x+1)+C(x1)=4x22x(x+1)(x1)

(x + 1)(x - 1)A + x(x-1)B + x(x + 1)C = 4x2 – 2

Ax2 – A + Bx2 – Bx + Cx2 + Cx = 4x2 – 2

(A + B + C)x2 + (C - B)x – A =  4x2 – 2

Comparing co-efficient of x2, x, and constants we get

A + B + C = 4 ……. (ii)

C – B = 0 

⇒ B = C

A = 2

∵ A + B + C = 4

⇒ 2 + B + C = 4

⇒ B + C = 2

∴ B = C = 1

Ax+B(x+1)+C(x1)=4x22x(x+1)(x1)=P

P=2x+1(x+1)+1(x1)

Pdx=(2x+1(x+1)+1(x1))dx

2dxx+dx(x+1)+dx(x1)

⇒ 2ln x + ln (x + 1) + ln (x - 1)

⇒ ln x2 + ln (x2 - 1)

⇒ ln x2(x2 - 1)

I.F.=ePdx

I.F =elnx2(x21)

⇒ I.F = x2 (x2 - 1)

Consider the initial value problem below. The value of y at x = In 2, (rounded off to 3 decimal places) is

dydx=2xy,y(0)=1

Answer (Detailed Solution Below) 0.8774 - 0.8952

Linear Question 7 Detailed Solution

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Concept:

The standard form of a first-order linear differential equation is,

dydx+Py=Q

Where P and Q are the functions of x.

Integrating factor, IF=ePdx

Now, the solution for the above differential equation is,

y(IF)=IF.Qdx

Calculation:

dydx=2xy,y(0)=1

dydx+y=2x

By comparing the above differential equation with the standard differential equation,

P = 1, Q = 2x

Integrating factor, IF=ePdx=e1dx=ex

Now, the solution is

y(ex)=ex(2x)dx

yex=2(xexex)+C

y=2(x1)+Cex

y(0) = 1

⇒ 1 = 2 (0 – 1) + C

⇒ C = 3

Now, the solution becomes

y = 2x – 2 + 3e-x

At x = ln 2,

y = 2 ln 2 – 2 + 3 (0.5) = 0.8862

Consider the differential equation (t281)dydt+5ty=sin(t) with y(1) = 2π. There exists a unique solution for this differential equation when t belongs to the interval

  1. (–2, 2)
  2. (–10, 10)
  3. (–10, 2)
  4. (0, 10)

Answer (Detailed Solution Below)

Option 1 : (–2, 2)

Linear Question 8 Detailed Solution

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(t281)dydt+5ty=sin(t)

dydt+5t(t281)y=sint(t281)

It is in the standard form of first order linear equation.

Integrating factor =e5t(t281)dt

=e52.2t(t281)dt

=e52ln(t281)=(t281)52

Solution of differential equation is:

y(t281)52=sint(t281).(t281)52dt+c

=sint(t281)32dt+c

If t = ±9,

If t = ±9 then the solution is not unique hence range (-10,10), (-10, 2), (0,10) can be eliminated, then left option is (-2,2)

The differential equation y=xdydx+x(dy/dx) is

  1. First-order and first-degree
  2. First-order and second-degree
  3. Second-order and first-degree
  4. Second-order and second-degree

Answer (Detailed Solution Below)

Option 2 : First-order and second-degree

Linear Question 9 Detailed Solution

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Concept:
Differential equation:

It is an equation that involves differential coefficients or differentials.

Order of differential equation:

It is the order of the highest derivative appearing in that equation.

Degree of the differential  equation:

It is the degree (power) of the highest derivative occurring in that equation after the equation has been expressed in a form, free from radicals and fractional powers as far as the derivatives are concerned.

Explanation:

Given D.E is

y=xdydx+x(dy/dx)

ydydx=x(dydx)2+x

Hence the highest order in the equation is 1 of dydxand the highest degree (power) of the equation is 2 of dydx

Hence the equation is First-order and Second-degree*

The integrating factor of equation sec2ydydx+xtany=x3 is

  1. ex22
  2. ex22
  3. ex2
  4. ex2

Answer (Detailed Solution Below)

Option 1 : ex22

Linear Question 10 Detailed Solution

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Concept:

The standard form of a first-order linear differential equation is,

Form 1: dydx+Py=Q

Where P and Q are the functions of x.

Integrating factor, IF=ePdx

Now, the solution for the above differential equation is,

y(IF)=IF.Qdx+C

Form 2: dxdy+Px=Q

Where P and Q are the functions of y.

Integrating factor, IF=ePdy

Now, the solution for the above differential equation is,

x(IF)=IF.Qdy+C

Calculation:

Given the differential equation is

sec2ydydx+xtany=x3

Put tan y = t

By differentiating with respect to t,

⇒ sec2 y dy = dt

Now, the given equation becomes

dtdx+xt=x3

Now, it is in the form of a first order linear differential equation.

Integrating factor =exdx=ex22

Which one of the following is the general solution of the first order differential equation

dydx=(x+y1)2, where x, y are real?

  1. y = 1 + x + tan–1 (x + c), where c is a constant.
  2. y = 1 + x + tan(x + c), where c is a constant.
  3. y = 1 – x + tan–1 (x + c), where c is a constant.
  4. y = 1 – x + tan(x + c), where c is a constant.

Answer (Detailed Solution Below)

Option 4 : y = 1 – x + tan(x + c), where c is a constant.

Linear Question 11 Detailed Solution

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Concept:

Variable separable

Equation of the form dydx=f(ax+by+c) can be reduced to variable separable form by putting ax + by + c = t

Calculation:

Given:

dydx=(x+y1)2

Substitute x + y – 1 = t

1+dydx=dtdx

dydx=dtdx1

dtdx1=t2

dtdx=1+t2

dt1+t2=dx

tan-1 t = x + c

t = x + y – 1

tan-1 (x + y – 1) = x + c

x + y – 1 = tan (x + c)

y = 1 – x + tan (x + c)

A curve passes through the point (x = 1, y = 0) and satisfies the differential equation dydx=x2+y22y+yx. The equation that describes the curve is

  1. ln(1+y2x2)=x1
  2. 12ln(1+y2x2)=x1
  3. ln(1+yx)=x1
  4. 12ln(1+yx)=x1

Answer (Detailed Solution Below)

Option 1 : ln(1+y2x2)=x1

Linear Question 12 Detailed Solution

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dydx=x2+y22y+yx

dydx=x22y+y2+yx

ydydx(1x+12)y2=x22

y2 = t

2ydyx=dtdx

12dtdx(1x+12)t=x22

dtdx2(1x+12)t=x2

The above is linear differential equation of the form

dtdx=+pt=Q

Integrating factor =e2(1x+12)dx

= e-2ln x. e-x

=elnx2.ex

=exx2

Solution is given by:

texx2=x2exx2dx+c

Substitute t = y2

y2exx2=ex+c

At, x = 1, y = 0

0 = -e-1 + c

c = e-1 = y­e

substituting in the solution

y2.exx2=ex+e1

(y2x2+1)ex=e1

(1+y2x2)=e1ex=ex1

Taking log both sides

ln(1+y2x2)=x1

A differential equation didt0.2i=0 is applicable over 10<t<10. If i(4)=10, then i(5) is _______.

Answer (Detailed Solution Below) 1.6 - 1.7

Linear Question 13 Detailed Solution

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Concept:

Linear Differential Equation: Linear differential equations are those in which the dependent variable, its derivatives occur only in the first degree and they are not multiplied together. It is of the form:

dnydxn+k1dn1ydxn1+k2dn2ydxn2++kny=0

Where, k1, k2, …kn are the constants.

The solution of the equation is given as:

y = C.F 

where C.F is the complementary function.

The above linear differential equation in the symbolic form is represented as

(Dn + k1 Dn-1 + k2 Dn-2 +…+ kn) y = 0

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Roots of Auxiliary Equation

Complementary Function

m1m2, m3, … (real and different roots)

C1em1x+C2em2x+C3em3x+

m1m1, m3, … (two real and equal roots)

(C1+C2x)em1x+C3em3x+

m1m1, 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 ± i β, α ± i β, m5, … (two pairs of equal imaginary roots)

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

 

Calculation:

didt0.2i=0(D0.2)i(t)=0D=0.2i(t)=k.e0.2t,10<t<10

At t=4;   10=K.e0.8

K=4.493i(5)=4.493×e1i(5)=1.65

The solution of xdydx+y=x4 with the condition y(1)=65 is

  1. y=x45+1x
  2. y=4x45+45x
  3. y=x45+1
  4. y=x55

Answer (Detailed Solution Below)

Option 1 : y=x45+1x

Linear Question 14 Detailed Solution

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Explanation:

xdydx+y=x4

dydx+yx=x3

This is single order differential equation of:

dydx+py=Q

then  P = 1/x and Q = x3

now IF=epdx=e1xdx=elnx=x

then solution of equation

⇒ y(IF)= ∫Q(IF)dx + C

y(x)=x3.xdx+cyx=x55+c

Applying boundary condition y(1) = 6/5

65(1)=(1)55+cC=1

then yx=x55+1y=x45+1x

The general solution of the differential equation dydx=1+cos2y1cos2x is

  1. tanycotx=c(cisaconstant)
  2. tanxcoty=c(cisaconstant)
  3. tany+cotx=c(cisaconstant)
  4. tanx+cotY=c(cisaconstant)

Answer (Detailed Solution Below)

Option 3 : tany+cotx=c(cisaconstant)

Linear Question 15 Detailed Solution

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dydx=1+cos2y1cos2xdydx=cos2ysin2xsec2ydy=cosec2x.dxtany=cotx+ctany+cotx=c
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