First order Differential Equation MCQ Quiz - Objective Question with Answer for First order Differential Equation - Download Free PDF

Last updated on Jul 7, 2025

Latest First order Differential Equation MCQ Objective Questions

First order Differential Equation Question 1:

If x = x(t), y = y(t) is the solution of the initial value problem dxdt=x4e2ty

dydt=e2txy

x(0) = 1, y(0) = 1

then which of the following statements are true? 

  1. limtt2x(t)y(t)=0
  2. x(1) = 0, y(12) = 0
  3. x(12) = 0, y (1) = 0
  4. limtt2x(t)y(t)=2

Answer (Detailed Solution Below)

Option :

First order Differential Equation Question 1 Detailed Solution

Concept:

Solving Coupled First-Order Linear Differential Equations:

  • The system of equations is: dxdt=x4e2ty and dydt=e2txy.
  • Since the coefficients contain exponential terms, use substitution techniques to simplify.
  • Consider the substitution y(t)=e2tz(t) to remove exponential terms in the second equation.

 

Calculation:

Substitute y(t)=e2tz(t):

dydt=2e2tz(t)+e2tz(t).

The second equation becomes:

2e2tz+e2tz=e2txe2tz

Cancel e2t terms:

z+3z=x    → (Equation 1).

Now substitute y=e2tz  into the first equation:

dxdt=x4e2t×e2tz=x4z

So, dxdtx=4z    → (Equation 2).

Differentiating Equation 2:

d2xdt2dxdt=4z.

From Equation 1, z=x3z:

d2xdt2dxdt=4x+12z.

But from Equation 2, z=xdxdt4:

So, d2xdt2dxdt=4x+12×xdxdt4.

That simplifies to:

d2xdt2dxdt=4x+3x3dxdt.

Simplify terms:

d2xdt2+2dxdtx=0.

This is a linear ODE whose general solution is:

x(t)=C1e(1+2)t+C2e(12)t.

Solving for y(t) using z(t) and initial conditions gives:

x(12)=0  , and y(1)=0 after evaluating constants correctly.

Also, the asymptotic term t2x(t)y(t) as t evaluates to 2 using dominant term analysis from the solution structure.

Final Answer Analysis:

Option 1: limtt2x(t)y(t)0 — False.

Option 2: Direct substitution shows x(1)0. — False.

Option 3: Verified from the solution, x(12)=0 and y(1)=0. —  Correct.

Option 4: Asymptotic analysis gives limtt2x(t)y(t)=2. — Correct.

∴ The correct answers are 3 and 4.

First order Differential Equation Question 2:

Let |dydx|+y2=2, y = y(x). Find the Number of solutions.

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

Answer (Detailed Solution Below)

Option 1 : 0

First order Differential Equation Question 2 Detailed Solution

Explanation:

|dydx|+y2=2

Let y(x) be any non-constant solution 

⇒ ∃ x ∈ R such that  y(x0) ≠ 0

i.e., |y(x0)| > 0

then at x0|dydx|+y2>0 contradiction

So it has no non-constant solution

If y(x) = a be a constant solution then |0|+a2=2

⇒ a2 = -2 ⇒ no such a exist.

So given ODE has no solution 

(1) is correct.

First order Differential Equation Question 3:

The solution of xlogxdydx+y=4logx is

  1. y = 4 log + c/log x : c is an arbitrary constant. 
  2. y = log + c/log x: c is an arbitrary constant. 
  3. y = 2log + c/log x : c is an arbitrary constant. 
  4. y = 4 log + c4logx : c is an arbitrary constant.

Answer (Detailed Solution Below)

Option 3 : y = 2log + c/log x : c is an arbitrary constant. 

First order Differential Equation Question 3 Detailed Solution

Explanation:  

xlogxdydx+y=4logx   

dividing both side by x log x

(dy /dx) + (y / (x log x)) = 4 / x

This is a first-order linear differential equation of the form:

(dy/dx) + P(x)y = Q(x)

where: P(x) = 1 / (x log x) and Q(x) = 4 / x

The integrating factor is given by :

I.F=eP(x)dx=e(1/(xlogx))dx   

u = log x   

du = (1/x) dx

Then:  ∫(1 / (x log x)) dx = ∫(1/u) du = ln |u| = ln |log x|

Therefore, the integrating factor is:

I.F=eln|logx|=|logx|   

Multiply the equation by the integrating factor:

|log x| (dy/dx) + (|log x| y / (x log x)) = 4|log x| / x

|log x| (dy/dx) + (y / x) = 4|log x| / x 

∫ d/dx (y |log x|) dx = ∫ 4|log x| / x dx 

y |log x| = 2(log x)² + C 

where C is the constant of integration

y = (2(log x)² + C) / |log x| 

Therefore, the solution to the given differential equation is:

y = 2 log x + C / |log x|  

Hence Option(3) is the correct answer.

First order Differential Equation Question 4:

The solution of the differential equation xdydx + y = x3y6 is : (where C is an arbitary constant)

  1. y5x5=52x2+C
  2. y5x2=52x2+C
  3. y5x5=52x5+C
  4. y2x5=52x2+c

Answer (Detailed Solution Below)

Option 1 : y5x5=52x2+C

First order Differential Equation Question 4 Detailed Solution

Concept:

 If  dydx+Py=Q 

where P and Q are functions of x only, then

it has ePdx
as an integrating factor and

its solution is given by

 y (I.F.)=∫Q(I.F.) dx+ C

Explanation: 

Given Differential equation is :

xdydx + y = x3y6 

⇒ dydx + 1x y = x3y6

Dividing both side by  y6 :

⇒ y6 dydx + 1x y5 =  x2 

Put  y5=u 

Then  dudx = (-5) y6 dydx

Now 15 dudx + 1xu = x2 

⇒ dudx - 5xu = -5 x2  

This is First order Differential equation: 

I.F= e5logx = 1x5

General solution will be : 

 u1x5 = 5x21x5dx + C

where C is an arbitrary constant.

Now Putting the value of u in terms of y:

 y5  1x5 = 51x3dx + C 

⇒ y5 x5 = 52 x2 + C 

Hence Option (1) is the correct answer.

First order Differential Equation Question 5:

Let y(x) be the solution of the differential equation

dydx=1+ysecx for x(π2,π2)

that satisfies y(0) = 0. Then, the value of y (π6) equals

  1. 3log(32)
  2. (32)log(32)
  3. (32)log3
  4. 3log3

Answer (Detailed Solution Below)

Option 1 : 3log(32)

First order Differential Equation Question 5 Detailed Solution

Explanation:

dydx=1+ysecx

dydxysecx=1

Which is first order linear differential equation

dydx+P(x)y=Q(x)

Where P(x)=secxandQ(x)=1

Integrating Factor (I.F.)=esecxdx=elog|secx+tanx|=1secx+tanx

Now solution of first order differential equation is;

y×I.F.=(Q×I.F.)dx+C

y(1secx+tanx)=1secx+tanxdx+C

⇒ ysecx+tanx=cosxsin2x+1dx+C

⇒ ysecx+tanx=log(1+sinx)+C

y(0)=0

0=log1+CC=0

⇒ y=(secx+tanx)log(1+sinx)

⇒ y(π6)=(secπ6+tanπ6)log(1+sinπ6)

⇒ y(π6)=(23+13)log(1+12)

⇒ y(π6)=3log(32)

Hence Option (1) is the correct answer.

Top First order Differential Equation MCQ Objective Questions

First order Differential Equation Question 6:

Suppose x : [0, ∞) → [0, ∞) is continuous and x(0) = 0. If (x(t)2 ≤ 2 + 0t x(s) ds, ∀t ≥ 0, then which of the following is TRUE?

  1. x(√2) ∈ [0, 2] 
  2. x(2)[0,32]
  3. x(2)[52,72]
  4. x(√2)[10, ∞)

Answer (Detailed Solution Below)

Option 1 : x(√2) ∈ [0, 2] 

First order Differential Equation Question 6 Detailed Solution

Explanation:

(x(t)2 ≤ 2 + 0t x(s) ds, ∀ t ≥ 0.....(i)

x : [0, ∞) → [0, ∞) is continuous and x(0) = 0

Let x(t) = t so it is continuous and x(0) = 0

So from (i) we get

t2 ≤ 2 + 0t s ds

⇒ t2 ≤ 2 + t2/2

⇒ t2/2 < 2

⇒ t< 4

⇒ - 2 < t < 2

⇒ - 2 < x(t) < 2

So (2), (3), (4) discard

(1) is correct 

First order Differential Equation Question 7:

Let |dydx|+|y|=0, y = y(x). Find the Number of solutions.

  1. 1
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 1 : 1

First order Differential Equation Question 7 Detailed Solution

Explanation:

|dydx|+|y|=0

Let y(x) be any non-constant solution 

⇒ ∃ x ∈ R such that

 y(x0) ≠ 0

i.e., |y(x0)| > 0

then at x0|dydx|+|y|>0 contradiction

So it has no non-constant solution

If y(x) = 0 then |dydx|+|y|=0

So given ODE has only trivial solution 

(1) is correct

First order Differential Equation Question 8:

Let |dydx|+y2=2, y = y(x). Find the Number of solutions.

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

Answer (Detailed Solution Below)

Option 1 : 0

First order Differential Equation Question 8 Detailed Solution

Explanation:

|dydx|+y2=2

Let y(x) be any non-constant solution 

⇒ ∃ x ∈ R such that  y(x0) ≠ 0

i.e., |y(x0)| > 0

then at x0|dydx|+y2>0 contradiction

So it has no non-constant solution

If y(x) = a be a constant solution then |0|+a2=2

⇒ a2 = -2 ⇒ no such a exist.

So given ODE has no solution 

(1) is correct.

First order Differential Equation Question 9:

Let |dydx|+|y|=1, y = y(x). Find the Number of solutions.

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

Answer (Detailed Solution Below)

Option 1 : 0

First order Differential Equation Question 9 Detailed Solution

Explanation:

|dydx|+|y|=1

Let y(x) be any non-constant solution 

⇒ ∃ x ∈ R such that  y(x0) ≠ 0

i.e., |y(x0)| > 0

then at x0|dydx|+|y|>0 contradiction

So it has no non-constant solution

If y(x) = a be a constant solution then |0|+|a|=1

⇒ |a| = -1 ⇒ no such a exist.

So given ODE has no solution 

(1) is correct.

First order Differential Equation Question 10:

Consider the ordinary differential equation y' = y(y - 1)(y - 2). Which of the following statements is true ? 

  1. If y(0) = 0.5 then y is decreasing
  2. If y(0) = 1.2 then y is increasing
  3. If y(0) = 2.5 then y is unbounded 
  4. If y(0) < 0 then y is bounded below

Answer (Detailed Solution Below)

Option 3 : If y(0) = 2.5 then y is unbounded 

First order Differential Equation Question 10 Detailed Solution

Concept:

A function y = f(x) is increasing if y' > 0 and decreasing if y' < 0

Explanation:

Given y' = y(y - 1)(y - 2)

critical points are given by

y' = 0

⇒ y(y - 1)(y - 2) = 0

⇒ y = 0, 1, 2

so we can see that

y' > 0 in the intervals (0, 1), (2, ∞) and so increasing

y' < 0 in the intervals (-∞, 0), (1, 2) and so decreasing 

(1): y(0) = 0.5, lies in the interval (0, 1) so y is increasing

(1) is false

(2): y(0) = 1.2, lies in the interval (1, 2) so y is decreasing

(2) is false

(3): y(0) = 2.5, lies in the interval (2, ) so y is increasing and hence unbounded

(3) is correct

(4): y(0) < 0, lies in the interval (-, 0) so y is decreasing so bounded above not bounded below

(4) is false

First order Differential Equation Question 11:

-x dy + y dx + e1/x dx = 0 then

  1. ∃ an I.F which is function of x only
  2.  any I.F which is function of y only
  3.  any I.F which is function of u, u = x. y only
  4.  any I.F which is function of u only, u=yx

Answer (Detailed Solution Below)

Option 1 : ∃ an I.F which is function of x only

First order Differential Equation Question 11 Detailed Solution

Concept:

An ODE Mdx + Ndy is said be exact if and only if My=Nx 

Explanation:

Given ODE

-x dy + y dx + e1/x dx = 0

⇒ (y + e1/x)dx - x dy = 0 

Comparing with the general form

M = y + e1/x, N = -x

⇒ My = 1, Nx = -1

As both are not equal so ODE is not exact.

Now, My - Nx = 2

(1): MyNxN=2x = f(x) 

⇒ IF = e2xdx=1x2 

(1) is true

(2): MyNxM=2y+e1/x = f(x,y), not function of y

⇒  any I.F. which is a function of y only.

(2) is false

(3): MyNxYNxM=2y(x)x(y+e1/x)=22xyxe1/xnot a function of u = x.y

⇒  any I.F which is a function of u-only

(3) is false

(4): MyNxYN+xM=2xe1/x

x2(MyNxYN+xM)=2xe1/x , not a function of yx

⇒  any I.F which is a function of u only, u = yx

(4) is false

First order Differential Equation Question 12:

Let y : ℝ → ℝ satisfy the initial value problem 

y'(t) = 1 - y2(t), t ∈ R, y(0) = 0 then  

  1. y(t1) = 1 for some t1 ∈ ℝ
  2. y(t) > -1 ∀ t ∈ ℝ
  3. y is strictly increasing in 
  4. y is increasing in (0, 1) and decreasing in (1, ∞) 

Answer (Detailed Solution Below)

Option :

First order Differential Equation Question 12 Detailed Solution

Explanation:

y'(t) = 1 - y2(t)

dydt=1y2

dy1y2=dt

Integrating both sides

12ln|1+y1y| = t + c

Given y(0) = 0

0 = 0 + c ⇒ c = 0

Hence  

12ln|1+y1y| = t

ln|1+y1y| = 2t

1+y1y = e2t

Using componendo and dividendo rule

22y=e2t+1e2t1

y=e2t1e2t+1

F2 Vinanti Teaching 09.08.23 D2

y(t) > -1 and y is strictly increasing in 

Options (2), (3) are correct

First order Differential Equation Question 13:

Consider the ODE ty˙3y=t2y12,y(1)=1. Find the value of y(2).

  1. 14
  2. 16
  3. 0
  4. 8

Answer (Detailed Solution Below)

Option 2 : 16

First order Differential Equation Question 13 Detailed Solution

Explanation:

ty˙3y=t2y12

dydt - 3yt = t√y

1ydydt -3yt = t...(i)

Let √y = x so 1ydydt = 2dxdt

then (i) becomes

2dxdt - 3tx = t which is linear

So IF = e32tdt = t3/2

Hence solution is

xt3/2 = 12∫ t3/2tdt + c

xt3/2 = √t + c

x = t2 + ct3/2

√y = t2 + ct3/2

Using y(1) = 1 we get c =0 or - 2

Bt when c = -2 then y = (t2 - 2t3/2)2 so y(1) = 0 which is contradiction

so y = t4

Hence y(2) = 24 = 16

(2) true

First order Differential Equation Question 14:

The singular solution of the differential equation

xyp2 - (x2 + y2 - 1)p + xy = 0, where p = dydx 

  1. y = 0
  2. y2 = (x - 1)2
  3. Does not exist
  4. None of these

Answer (Detailed Solution Below)

Option 2 : y2 = (x - 1)2

First order Differential Equation Question 14 Detailed Solution

Explanation:

xyp2 - (x2 + y2 - 1)p + xy = 0....(i) where p = dydx 

Let x2 = u and y2 = v, P = dudv and Q = dvdu

So Q = pyx ⇒ p = Qxy

Then putting the value of p in (i) we get

xyQx2y2 - (x+ y2 - 1)Qxy + xy = 0

Multiplying both side by y/x we get

x2Q2 - (x+ y2 - 1)Q + y2 = 0

uQ2 - (u + v - 1)Q + v = 0

Factorizing we get

(Q - 1)(uQ - v) = - Q

uQ - v = - QQ1

v = uQ - QQ1....(ii)

Differentiating with respect to u

Q = Q + udQdu + dQdu(Q1)2

dQdu(u + 1(Q1)2) = 0

⇒ dQdu = 0 ⇒ Q = c and

u = - 1(Q1)2 ⇒ u = - 1(c1)2 ⇒ c = 1 + 1u ⇒ c = 1 +1x

Hence (ii) ⇒ v = uc - cc1

Replacing c by  1 +1x, u = x2, v = y2 we get the singular solution as 

y2 = x2 + 2x + 1

⇒ y2 = (x - 1)2

(2) correct

First order Differential Equation Question 15:

Let f : ℝ2 + 2 be a nonzero smooth vector field satisfying divf ≠ 0. Which of the following are necessarily true for the ODE x˙ = f(x)?

  1. There are no equilibrium points
  2. There are no periodic solutions
  3. All the solutions are bounded
  4. All the solutions are unbounded

Answer (Detailed Solution Below)

Option :

First order Differential Equation Question 15 Detailed Solution

Explanation:

Recall: div f = f=(i^x+j^y)(f1i^+f2j^)

f1x+f2y

In ℝ2, if div f ≠ 0 ⇒ there is no fixed point.

⇒ there is no equilibrium point.

Consider, f(x, y) = exî + eyĵ then div f ≠ 0 and smooth vector field.

Here, f is not periodic ⇒ there is no periodic solution.

Also, f is not bounded ⇒ solution is not bounded.

Method-II

Given, f ≠ 0, div f ≠ 0 ⇒ k(x) ≠ 0, ∀ x ∈ R

Now, for equilibrium point, x = 0 ⇒ f(x) = 0 which is contradiction as f(x) ≠ 0 (given)

⇒ There are no equilibrium point 

Option (1) is correct.

If there are periodic solutions then ∃ x0 ∈ ℝ s. t. f(x0) = 0 but f(x) ≠ 0 ∀ x ∈ R ⇒ there are no periodic solution.

option (2) is correct.

Recall: periodic ⇒ bounded.

∴ Solutions are unbounded.

Can't be bounded.

option (3) is not correct and option (4) is correct.

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