Initial Value Problem MCQ Quiz - Objective Question with Answer for Initial Value Problem - Download Free PDF

Last updated on Jun 26, 2025

Latest Initial Value Problem MCQ Objective Questions

Initial Value Problem Question 1:

The initial value problem y' = √y, y(0) = 0 has  

  1. unique solution
  2. two solution
  3. no solution
  4. infinitely many solution

Answer (Detailed Solution Below)

Option 4 : infinitely many solution

Initial Value Problem Question 1 Detailed Solution

Concept:

Let, dydx = ayα, y(β) = 0, α ∈ (0, 1), be an initial value problem then this IVP has 
 
(i) unique solution if a < 0 and
 
(ii) infinitely many solutions if a > 0

 

Explanation:

y' = √y, y(0) = 0 ....(i)

Comparing (i) with the above original equation we get

a = 1 > 0

So, the given ODE has infinitely many solutions.

Option (4) is true.

Initial Value Problem Question 2:

Given ODE

 dxdt = cos x,  x(0) = 1.

Which of the following statements is not true?

  1. There might not be any solution around 0
  2. There is at most one local solution at t = 0
  3. There always exists a global solution defined in [0, ∞)
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : There might not be any solution around 0

Initial Value Problem Question 2 Detailed Solution

Concept:

Lipschitz Condition: A function f(x, y) is said to satisfy a Lipschitz condition in a region R if there exists a positive real number k such that

|fy| ≤ k

Explanation:

Given ODE is

dxdt = cos x,  x(0) = 1

Here f(x, t) = cos x

fx = - sin x

|fx| = |sin x| ≤ 1

So, f satisfying Lipschitz condition

Hence the ODE has unique solution in around t = 0.

Hence there is at most one local solution at t = 0

Therefore (4) is true.

Also, cos x is a global Lipschitz so always exists global solution.

There always exists a global solution defined in [0, ∞).

Hence (3) is true.

(1) is not true.

Initial Value Problem Question 3:

Find the largest interval for which the IVP dydx=ey,y(0)=0 has unique solution

  1. |x| < ∞ 
  2. |x| < 1
  3. |x| < e
  4. |x| < e-1

Answer (Detailed Solution Below)

Option 4 : |x| < e-1

Initial Value Problem Question 3 Detailed Solution

Concept:
 
Let us consider an initial value problem (IVP)
 
dydx = f(x,y); y(x0) = y0 ……(1)
 
and R = {(x,y):│x - x0| ≤ a, |y - y0| ≤ b} 
 
which satisfy the following conditions
 
f(x, y) is bounded in R i.e., there exist real number M such that |f(x,y)| ≤ M
 
and f(x, y) is continuous in R
 
then IVP (1) has a solution in the interval |x - x0| ≤ h where h = min⁡(a, b/M).
 
Explanation: 
 
dydx=ey
 
⇒ e-y dy = dx
 
⇒ -e-y = x + c
 
Using y(0) = 0 we get
⇒ - 1 = c
 
So, we get
 - e-y = x - 1
⇒ e-y =1-x
⇒ y = log⁡(1-x)
 
Which defined on
1 - x > 0 ⇒ x < 1 ⇒ -∞ < x < 1
 
Consider the interval |x - x0| ≤ a, |y - y0| ≤ b ⇒ |x|<∞, |y| ≤ b
 
Now, f(x,y) is continuous and |f(x,y)| ≤ eb i.e., f is bounded
 
Also, |fy| = |ey| ≤ eb, satisfy Lipschitz condition.
 
Then there exist an unique solution in the interval |x| ≤ h where h = min (1, b/M) = min⁡(∞, b/eb)
 
Let, ϕ(b) = b/eb = be-b
ϕ'(b) = e-b - be-b = e-b (1 - b)
 
Critical point is
ϕ'(b) = 0 ⇒ b = 1
 
Now, ϕ''(b) = -e-b(1 - b) -e-b
ϕ''(1) = -e-1 < 0, maximum value exits.
 
Hence max⁡ ϕ(b) = e-1
 
Therefore, min⁡(∞,b/eb) = e-1
 
Hence required interval is |x| < e-1.
 
Option (4) is true.

Initial Value Problem Question 4:

The initial value problem dydxyx=x, x > 0; y(0) = 1 has

  1. Infinitely many solutions.
  2. Exactly two solutions.
  3. A unique solution.
  4. No solution.

Answer (Detailed Solution Below)

Option 4 : No solution.

Initial Value Problem Question 4 Detailed Solution

Explanation:

dydxyx=x which is linear in y

IF = e1xdx = e-logx = 1/x

So, general solution is

yx=x.1xdx + c

 ⇒ yx = x + c

⇒ y = x2 + cx

Using y(0) = 1 

⇒ 1 = 0 which is not possible.

Hence, the differential equation has no solution.

(4) is true.

Initial Value Problem Question 5:

The differential equation

{dydt=y2kk+2 for t>0,y(0)=0
has infinitely many solution if

  1. 0 < k < 3/2
  2. 1 < k < 3
  3. 0 < k < 1
  4. 0 < k < 2

Answer (Detailed Solution Below)

Option 4 : 0 < k < 2

Initial Value Problem Question 5 Detailed Solution

Concept:

If dydx=yα, α ∈ (0, 1) and y(a) = 0, a ∈ R then the differential equation has infinite number of independent solution.

Explanation:

Here dydt=y2kk+2, y(0) = 0 has infinitely many solution is

0 < 2kk+2 < 1

⇒ k > 0 and 2k < k + 2, k ≠ -2

⇒ k < 2

Hence, 0 < k < 2

So, the differential equation has infinitely many solutions on (0, ∞) if 0 < k < 2.

Option (4) is correct

Top Initial Value Problem MCQ Objective Questions

Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem

ẋ = f(t, x), x(t0) = x0

for (t0, x0) ∈ ℝ2. Suppose that J(t0, x0) represents the maximal interval of existence for the initial value problem. Which of the following statements is true?

  1. J(t0, x0) = ℝ.
  2. J(t0, x0) is an open set.
  3. J(t0, x0) is a closed set.
  4. J(t0, x0) could be an empty set.

Answer (Detailed Solution Below)

Option 2 : J(t0, x0) is an open set.

Initial Value Problem Question 6 Detailed Solution

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Explanation:

Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem

ẋ = f(t, x), x(t0) = x0

for (t0, x0) ∈ ℝ2

By using Picard's theorem we know that solutions lie in the interval 

|xxo|<a;|tto|<b

which is an open interval 

Therefore, the Correct Option is Option (2).

Let y0 > 0, z0 > 0 and α > 1.

Consider the following two differential equations:
(){dydt=yα for t>0,y(0)=y0(){dzdt=zα for t>0,z(0)=z0
We say that the solution to a differential equation exists globally if it exists for all t > 0.

Which of the following statements is true?

  1. Both (*) and (**) have global solutions

  2. None of (*) and (**) have global solutions
  3. There exists a global solution for (*) and there exists a T < ∞  such that limtT|z(t)|=+
  4. There exists a global solution for (**) and there exists a T < ∞ such that limtT|y(t)|=+

Answer (Detailed Solution Below)

Option 4 : There exists a global solution for (**) and there exists a T < ∞ such that limtT|y(t)|=+

Initial Value Problem Question 7 Detailed Solution

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Explanation:

 dydt=yα

y1α1α=t+c

and y(0) = y0 ⇒ y01α1α=c

then y1-α = (1 - α)t + y01-α

and If α > 1  1 - α < 0. suppose 1 - α = -a, a > 0

then (1) y-a = -at + y0-a 

⇒ ya=1y0aat

then for y0-a - at = 0, solution does not exist.

t=yoaa=1ayoa>o

(∵ y0 > 0, a > 0) 

∴ (*) has no global solution

as 1ay0a ∈ (0, ∞)

options (1) and (3) are false

(* *) dzdt=zα ⇒ z1α1α=t+c

and z0 = z(0)  z01α1α=c

∴ z1-α = -(1 - α)t + z01-α ....(ii)

and for α > 1 ⇒ 1 - α < 0. So, Suppose 1 - α = - b, b > 0

then (ii) ⇒ z-b = bt + zo-b  zb=1bt+z0b

and for bt + z0-b​ = 0 solution does not exist 

⇒ t=z0bb=1z0bb < o

(∵ zo > o, b > 0)

So, ∀ t > 0, solution exist of (* *)

 (**) have global solutions.

option (2) is false.

Also, for T < ∞, take T = 1ay0a

limtT|y(t)|=limtT|(1y0aat)1/a|

=limt1ay0a|(y01aat)1/a|

= + ∞ 

 option (4) is true.

Consider the initial value problem (IVP) 

{y(x)=|y(x)+ε|,xR y(0)=y0

Consider the following statements: 

S1: There is an ε > 0 such that for all y0 ∈ ℝ, the IVP has more than one solution.

S2: There is a y0R such that for all ε > 0, the IVP has more than one solution.  

Then

  1. both S1 and S2 are true
  2. S1 is true but S2 is false
  3. S1 is false but S2 is true
  4. both S1 and S2​ are false

Answer (Detailed Solution Below)

Option 4 : both S1 and S2​ are false

Initial Value Problem Question 8 Detailed Solution

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Concept:

 Lipschitz condition:  A function f(x,y)  satisfies the Lipschitz condition with respect to y in a domain

DR2 if there exists a constant L>0 such that for any (x,y1) and (x,y2) in  D , the following inequality holds

 |f(x,y1)f(x,y2)|L|y1y2|.

The constant L is called the Lipschitz constant.

Explanation:

y(x)=|y(x)|+ϵ,xR

y(0)=y0, where ϵ>0 is a constant, and y0R .

S1: There is an ϵ>0 such that for all y0R , the IVP has more than one solution.
 

S2: There is a y0R  such that for all ϵ>0 , the IVP has more than one solution.

The differential equation is y(x)=|y(x)|+ϵ . Since ϵ>0 , the right-hand side of the

equation is always positive and well-defined for any y0R .

In general, the uniqueness of solutions to IVPs can often be determined by the Lipschitz condition.

For a functionf(y)=|y|+ϵ , we need to check if the function satisfies the Lipschitz condition with respect to y .

f(y)=12|y|+ϵ.

This function is continuous and bounded for all y0R because ϵ>0 ensures no singularity at y = 0 .

Therefore, the function satisfies the Lipschitz condition, ensuring that the IVP has a unique solution for each y0R when ϵ>0 .

S1: This statement claims that there is some ϵ>0 for which the IVP has more than one solution for all y0R .

From our uniqueness analysis (Lipschitz condition), the IVP actually has a unique solution for all ϵ>0 and y0R .

Therefore, S1 is false.

S2: This statement claims that for some y0R and for all ϵ>0, the IVP has more than one solution.

Based on the same reasoning (Lipschitz condition and continuity of the derivative), the solution remains

unique for all ϵ>0 and for any y0 . Therefore, S2 is also false.

Both statements S1 and S2 are false.

Thus, the correct option is 4).

Let k be a positive integer. Consider the differential equation

{dydt=y5k5k+2 for t>0,y(0)=0
Which of the following statements is true?

  1. It has a unique solution which is continuously differentiable on (0, ∞)
  2. It has at most two solutions which are continuously differentiable on (0, ∞).
  3. It has infinitely many solutions which are continuously differentiable on (0, ∞).

  4. It has no continuously differentiable solution on (0, ∞)

Answer (Detailed Solution Below)

Option 3 :

It has infinitely many solutions which are continuously differentiable on (0, ∞).

Initial Value Problem Question 9 Detailed Solution

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Concept:

If dydx=yn, n ∈ (0, 1) and y(a) = 0, a ∈ R then the differential equation has infinite number of independent solution.

Explanation:

Here dydt=y5k5k+2, k ∈ N, y(0) = 0

∵ n =  5k5k+2 < 1 ∀ k ∈ N

Hence It has infinitely many solutions which are continuously differentiable on (0, ∞).

Option (3) is correct

Initial Value Problem Question 10:

Initial value problem, xdydx=y, y(0) = 0, x > 0

  1. there will be no solution
  2. Unique solution
  3. only and only two solutions
  4. infinite solution

Answer (Detailed Solution Below)

Option 4 : infinite solution

Initial Value Problem Question 10 Detailed Solution

Concept:

A variable separable differential equation is a type of first-order ordinary differential equation : dydx=f(x)g(y), Where f(x) & g(y) are function wrt x & y.

Explanation:

xdydx=y,  y(0) = 0, x > 0

⇒ dydx=yx

⇒ dyy=dxx

Now, Integrate both sides

logy = logx + logc

⇒ y=cx

Passes through (0,0)

⇒ 0=0

⇒ Infinite solution

Initial Value Problem Question 11:

Consider the following two initial value ODEs

(A) dxdt=x3,x(0)=1;

(B) dxdt=xsinx2,x(0)=2.

Related to these ODEs, we make the following assertions.

I. The solution to (A) blows up in finite time.

II. The solution to (B) blows up in finite time.

Which of the following statements is true?

  1. Both (I) and (II) are true
  2. (I) is true but (II) is false
  3. Both (I) and (II) are false
  4. (I) is false but (II) is true

Answer (Detailed Solution Below)

Option 2 : (I) is true but (II) is false

Initial Value Problem Question 11 Detailed Solution

Calculation:

(A) dxdt=x3dxx3=dt

Integrating both sides

12x2=t+c

Since given x(0) = 1 ⇒ 12 = 0 + c ⇒ = 12

So

12x2=t12

1x2=2t+1

x2=112t

x=±112t

Hence |x(t)| → ∞ at t=12

Therefore, solution (A) blows up in a finite time.

∴ Statement (I) is true.

(B) dxdt=xsin(x2)

Here |dxdt|=|xsinx2||x|

|x| which is the slope of the function x doesn't blow up rapidly, It increases slowly.

So x does not blow up in finite time.

Hence the solution of (B) does not blow up in finite time.

∴ Statement (II) is false.

The correct answer is option 2.

Initial Value Problem Question 12:

Consider the initial value problem

dydx+αy=0,

y(0) = 1, 

where α ∈ ℝ. Then

  1. there is an α such that y(1) = 0
  2. there is a unique α such that limx y(x) = 0
  3. there is NO α such that y(2) = 1
  4. there is a unique α such that y(1) = 2

Answer (Detailed Solution Below)

Option 4 : there is a unique α such that y(1) = 2

Initial Value Problem Question 12 Detailed Solution

Solution: 

Given differential equation 

dydx+αy = 0, y(0) = 1 where αR

dydx=αy

by using variable separable form 

dyy=αdx

Integrate, both sides 

log y = - αx + log c1

y = c1eαx  

Given initial condition y(1) = 0 then, c1=1 

So, the solution is

y = eαx

(1): y(1) = eα ≠ 0 for any α

So there does not exist an α such that y(1) = 0

(1) is false

(2): limxy(x) = limxeαx = 0 for all α > 0

So, there is no unique α such that limx y(x) = 0

(2) is false

(3): y(2) = 1

⇒ e2α = 1 ⇒ α = 0 

There is a α such that y(2) = 1

(3) is false

(4): y(1) = 2

⇒ eα = 2 ⇒ - α = log(2) ⇒ α = - log(2) 

Hence there is a unique α such that y(1) = 2

(4) is correct

Initial Value Problem Question 13:

Let us consider the following two initial value problems

(P) {x(t)=x(t),t>0,x(0)=0,

and

(Q) {y(t)=y(t),t>0,y(0)=0.

Which of the following statements are true?

  1. (P) has a unique solution in [0, ∞).
  2. (Q) has a unique solution in [0, ∞).
  3. (P) has infinitely many solutions in [0, ∞).
  4. (Q) has infinitely many solutions in [0, ∞).

Answer (Detailed Solution Below)

Option :

Initial Value Problem Question 13 Detailed Solution

Concept:

The differential equation y' = kyα, y(a) = 0 where α ∈ (0, 1) and a ∈ R has 

(i) unique solution if k < 0 and

(ii) infinitely many solution if k > 0

Explanation:

(P) {x(t)=x(t),t>0x(0)=0 then x'(t) = x1/2, x(0) = 0 

Here α = 1/2 ∈ (0, 1) and k = 1 > 0

So (P) has infinitely many solutions in [0, ∞).

Option (3) is correct

(Q) {y(t)=y(t),t>0y(0)=0 then y'(t) = y1/2, y(0) = 0

here α = 1/2 ∈ (0, 1) and k = - 1 < 0 

So (Q) has a unique solution in [0, ∞).

Option (2) is correct

Initial Value Problem Question 14:

Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem

ẋ = f(t, x), x(t0) = x0

for (t0, x0) ∈ ℝ2. Suppose that J(t0, x0) represents the maximal interval of existence for the initial value problem. Which of the following statements is true?

  1. J(t0, x0) = ℝ.
  2. J(t0, x0) is an open set.
  3. J(t0, x0) is a closed set.
  4. J(t0, x0) could be an empty set.

Answer (Detailed Solution Below)

Option 2 : J(t0, x0) is an open set.

Initial Value Problem Question 14 Detailed Solution

Explanation:

Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem

ẋ = f(t, x), x(t0) = x0

for (t0, x0) ∈ ℝ2

By using Picard's theorem we know that solutions lie in the interval 

|xxo|<a;|tto|<b

which is an open interval 

Therefore, the Correct Option is Option (2).

Initial Value Problem Question 15:

Given ODE

 dxdt = cos x,  x(0) = 1.

Which of the following statements is not true?

  1. There might not be any solution around 0
  2. There is at most one local solution at t = 0
  3. There always exists a global solution defined in [0, ∞)
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : There might not be any solution around 0

Initial Value Problem Question 15 Detailed Solution

Concept:

Lipschitz Condition: A function f(x, y) is said to satisfy a Lipschitz condition in a region R if there exists a positive real number k such that

|fy| ≤ k

Explanation:

Given ODE is

dxdt = cos x,  x(0) = 1

Here f(x, t) = cos x

fx = - sin x

|fx| = |sin x| ≤ 1

So, f satisfying Lipschitz condition

Hence the ODE has unique solution in around t = 0.

Hence there is at most one local solution at t = 0

Therefore (4) is true.

Also, cos x is a global Lipschitz so always exists global solution.

There always exists a global solution defined in [0, ∞).

Hence (3) is true.

(1) is not true.

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