Initial Value Problem MCQ Quiz in मराठी - Objective Question with Answer for Initial Value Problem - मोफत PDF डाउनलोड करा

Last updated on Mar 20, 2025

पाईये Initial Value Problem उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Initial Value Problem एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Initial Value Problem MCQ Objective Questions

Top Initial Value Problem MCQ Objective Questions

Initial Value Problem Question 1:

Consider the initial value problem (IVP) y(x)=sin(y(x))1+y4(x), x ∈ ℝ, y(0) = y0

Then which of the following statements are true? 

  1. There is a positive y0 such that the solution of the IVP is unbounded 
  2. There is a negative y0 such that the solution of the IVP is bounded 
  3. For every y0 ∈ ℝ, every solution of the IVP is bounded 
  4. For every y0 ∈ ℝ, there is a solution to the IVP for all x ∈ ℝ  

Answer (Detailed Solution Below)

Option :

Initial Value Problem Question 1 Detailed Solution

Concept:

Existence of Solutions: The right-hand side sin(y(x))1+y(x)4. is smooth and well-behaved for all yR.

This ensures the existence of a global solution for any initial condition y0R , and the solution exists for all xR .

Explanation:

We are given the initial value problem (IVP)

y(x)=sin(y(x))1+y(x)4,xR,y(0)=y0.

The right-hand side of the differential equation is:

sin(y(x))1+y(x)4.

Where numerator is bounded between -1 and 1 for all values of y(x) .

The denominator 1+y(x)4 grows rapidly as |y(x)| increases, which means that the right-hand 

side sin(y(x))1+y(x)4. approaches 0 as |y(x)| . Therefore, the growth rate of y(x) is controlled by the

denominator, and the differential equation indicates that large values of y(x) will not grow rapidly

because the right-hand side becomes small as |y(x)| increases.

Given that the right-hand side tends to zero for large |y(x)| , this suggests that the solutions will be bounded for all

initial conditions y0 . Even if y(x) starts at a large value, the small right-hand side means that y(x) will not grow indefinitely.

For both positive and negative initial values  y0 , the function y'(x) will be bounded because the numerator

sin(y(x)) is bounded and the denominator 1+y(x)4 grows rapidly for large |y(x)| . Hence, all

solutions are bounded for every initial value y0 .

There is no initial value y0 that leads to an unbounded solution.

Since the function on the right-hand side sin(y(x))1+y(x)4. is smooth and well-behaved for all yR , there is a

unique solution to the IVP for any initial condition y0R , and the solution exists for all xR .

Option 1: There is a positive y0 such that the solution of the IVP is unbounded. False — all solutions are bounded, regardless of the initial value y0
  
Option 2: There is a negative y0 such that the solution of the IVP is bounded. True — all solutions are bounded, including those with negative initial values.
  
Option 3: For every y0R , every solution of the IVP is bounded. True — all solutions are bounded for any initial value y0
  
Option 4: For every y0R , there is a solution to the IVP for all xR . True — the IVP has a global solution for any

initial condition because the right-hand side is smooth and well-behaved.

Hence, correct options are 2), 3) and 4).

Initial Value Problem Question 2:

Consider the problem

y' = (1 - y2)10 cos y, y(0) = 0.

Let J be the maximal interval of existence and K be the range of the solution of the above problem. Then which of the following statements are true? 

  1. J = ℝ
  2. K = (-1, 1)
  3. J = (-1, 1)
  4. K = [-1, 1]

Answer (Detailed Solution Below)

Option :

Initial Value Problem Question 2 Detailed Solution

Concept -

(1) If f is bounded and continuously differentiable on R then maximum interval of existence is R.

Explanation -

We have y' = (1 - y2)10 cos y, y(0) = 0.

Now f(x,y) = (1 - y2)10  cos y

Clearly f is continuous and differentiable function.

Now |f| ≤ (1 - y2)10 ≤ M  ∀ y ∈ R

Hence f is bounded as well

Therefore maximum interval of existence is R.

Now the function f = (1 - y2)10  cos y is even function as (1 - y2)10 is even and  cos y is also even function.

⇒ y' = even

⇒ y = odd 

Let assume y = x is odd function.

Now 1 = (1 - x2)10  cos x

Now according to the options, we take x = 1

then equation is not satisfied 1 ≠ 0

Hence the range of the solution (-1,1)

Initial Value Problem Question 3:

Consider the following initial value problem

y=y+12|sin(y2)|,, x > 0, y(0) = -1

Which of the following statements are true? 

  1. there exists an α ∈ (0, ∞) such that limxα|y(x)|=
  2. y(x) exists on (0, ∞) and it is monotone  
  3. y(x) exists on (0, ∞), but not bounded below
  4. y(x) exists on (0, ∞), but not bounded above

Answer (Detailed Solution Below)

Option :

Initial Value Problem Question 3 Detailed Solution

Explanation -

We have y=y+12|sin(y2)|,, x > 0, y(0) = -1

If a solution exist for the differential equation then Lipchitz condition is satisfied.

Now yy=12|sin(y2)|12

⇒ yy=12

Now the solution of the differential equation is -

C.F. = C1ex

PI = 1D1.12=12

Hence y = C1ex - 1/2

Now use the initial condition y(0) = -1 ⇒ C1 = -1/2

⇒ y = -1/2 ex - 1/2

⇒ y = -1/2 ex < 0 

⇒ y is monotone decreasing.

Now limxα|y(x)|= ⇒ 1/2 ex + 1/2 → ∞ ⇒ x  → ∞

but there does not exists an α ∈ (0, ∞)

Now as x → ∞ ,  y = -1/2 ex - 1/2 → - ∞ and y is decreasing as well.

So clearly it is not bounded below but it is bounded above.

Initial Value Problem Question 4:

Consider the initial value problem

dydx=f(x,y), y(x0) = y0

where f is a twice continuously differentiable function on a rectangle containing the point (x0, y0). With the step-size h, let the first iterate of a second order scheme to approximate the solution of the above initial value problem be given by

y= y+ Pk+ Qk2,

where k= hf(x0, y0), k= hf(x+ α0h, y+ β0k1) and P, Q, α0, β0 ∈ ℝ.

Which of the following statements are correct? 

  1. If α0 = 2, then β0 = 2, P=34,Q=14
  2. If β0 = 3, then α0 = 3, P=56,Q=16
  3. If α0 = 2, then β0 = 2, P=14,Q=34
  4. If β0 = 3, then α0 = 3, P=16,Q=56

Answer (Detailed Solution Below)

Option :

Initial Value Problem Question 4 Detailed Solution

Explanation:

the initial value problem

dydx=f(x,y), y(x0) = y0

where f is a twice continuously differentiable function on a rectangle containing the point (x0, y0). With the step-size h, let the first iterate of a second order scheme to approximate the solution of the above initial value problem be given by

y= y+ Pk+ Qk2,

where k= hf(x0, y0), k= hf(x+ α0h, y+ β0k1) and P, Q, α0, β0 ∈ ℝ.

Compared with the Explicit Runge–Kutta method we get the relation,

P + Q = 1....(i)

Qα=1/2....(ii)

Qβ= 1/2....(iii)

Option (1): 

If α0 = 2, then by (ii) Q = 1/4

Then by (iii), β0 = 2

Hence by (i) P = 3/4

Therefore we get

If α0 = 2, then β0 = 2, P=34,Q=14

Option (1) is true, (3) is false

Option (2): 

If β0 = 2, then by (iii) Q = 1/6

Then by (ii), α0 = 3

Hence by (i) P = 5/6

Therefore we get

If β0 = 3, then α0 = 3, P=56,Q=16

Option (2) is true, (3) is false

Initial Value Problem Question 5:

Let y be a solution of

(1 + x2)y" + (1 + 4x2)y = 0, x > 0

y(0) = 0. Then y has 

  1. infinitely many zeros in [0, 1]
  2. infinitely many zeros in [1, ∞)
  3. at least n zeros in [0, nπ], n ∈ ℕ  
  4. at most 3n zeros in [0, nπ], n ∈ ℕ

Answer (Detailed Solution Below)

Option :

Initial Value Problem Question 5 Detailed Solution

Concept:

Let u(x) be a non-trivial solution to the ODE u'' + q(x)u = 0 where q(x) > 0 for all x > 0. If 1q(x)dx = ∞ the u(x) has infinitely many zeros on the positive x-axis.

(ii) Let u(x) be a non-trivial solution to the ODE u'' + q(x)u = 0 on a closed interval [a, b] then u(x) has at most a finite number of zeros on [a, b]  

(iii) Let ϕ (x) and ψ(x) be two non-trivial solutions of y'' + p(x)y = 0, y'' + q(x)y = 0 respectively on an interval I where p(x) ≥ q(x) and p(x) ≠ q(x). Then between any two zeros of ψ, there is at least one zero of ϕ.  

Explanation:

(1 + x2)y" + (1 + 4x2)y = 0, x > 0

⇒ y'' + 1+4x21+x2y = 0

q(x) = 1+4x21+x2
now,

 1q(x)dx=01+4x21+x2dx

=1(431+x2)dx=[4x3tan1x]0=

Option (2) is correct.

By (ii), option (1) is not correct and option (4) is not correct.

Given equation

(1 + x2)y" + (1 + 4x2)y = 0, x > 0...(a)

Let y'' + y = 0....(b)

So here p(x) ≥ q(x) for all x as 1+4x21+x2 ≥ 1

(b) has a solution sin(x) which has zeros at points x = 0, kπ, for k ∈  

(a) has at least n zeros in [0, nπ], n ∈ ℕ 

Option (3) is correct  

Initial Value Problem Question 6:

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 6 Detailed Solution

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 7:

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 7 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 8:

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 8 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

Initial Value Problem Question 9:

let us consider an IVP dxdt + (sin t) x = t, x(0) = 1. then x(1) is

  1. 2 + ecos1
  2. 1 + ecos1
  3. 1 + sin 1
  4. 3 + ecos1

Answer (Detailed Solution Below)

Option 2 : 1 + ecos1

Initial Value Problem Question 9 Detailed Solution

Explanation:

dxdt + (sin t) x = t, x(0) = 1

which is linear in x

IF  = esintdt=ecost

So, solution is

xecost = ∫ tecosttdt + c

x = ecost∫ tecostdt + cecost

x(0) = 1 ⇒ c = 1

hence

x(t) = ecosttecostdt + ecost

Therefore

x(1) = ecos1 ecos1 + ecos1 = 1 + ecos1

Option (2) is true.

Initial Value Problem Question 10:

Consider the initial value problem

dxdt=x2,x(1)=1;  

Which of the following statements is true?

  1. The solution blows up in finite time.
  2. x(t) is finite for all t
  3. x(t) → ∞ as t → ∞
  4. none of the above

Answer (Detailed Solution Below)

Option 1 : The solution blows up in finite time.

Initial Value Problem Question 10 Detailed Solution

Calculation:

dxdt=x2dxx2=dt

Integrating both sides

1x=t+c

Since given x(1) = 1 ⇒ -1 = 1 + c ⇒ c = -2

So

1x=t2

⇒ x = 12t

As, x(t) → ∞ at t = 2

Therefore, solution blows up in a finite time.

The correct answer is option 1.

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