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
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)
Then which of the following statements are true?
Answer (Detailed Solution Below)
Initial Value Problem Question 1 Detailed Solution
Concept:
Existence of Solutions: The right-hand side
This ensures the existence of a global solution for any initial condition
Explanation:
We are given the initial value problem (IVP)
The right-hand side of the differential equation is:
Where numerator is bounded between -1 and 1 for all values of y(x) .
The denominator
side
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
For both positive and negative initial values
solutions are bounded for every initial value
There is no initial value
Since the function on the right-hand side
unique solution to the IVP for any initial condition
Option 1: There is a positive
Option 2: There is a negative
Option 3: For every
Option 4: For every
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?
Answer (Detailed Solution Below)
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
Which of the following statements are true?
Answer (Detailed Solution Below)
Initial Value Problem Question 3 Detailed Solution
Explanation -
We have
If a solution exist for the differential equation then Lipchitz condition is satisfied.
Now
⇒
Now the solution of the differential equation is -
C.F. = C1ex
PI =
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
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
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
y1 = y0 + Pk1 + Qk2,
where k1 = hf(x0, y0), k2 = hf(x0 + α0h, y0 + β0k1) and P, Q, α0, β0 ∈ ℝ.
Which of the following statements are correct?
Answer (Detailed Solution Below)
Initial Value Problem Question 4 Detailed Solution
Explanation:
the initial value problem
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
y1 = y0 + Pk1 + Qk2,
where k1 = hf(x0, y0), k2 = hf(x0 + α0h, y0 + β0k1) and P, Q, α0, β0 ∈ ℝ.
Compared with the Explicit Runge–Kutta method we get the relation,
P + Q = 1....(i)
Qα0 =1/2....(ii)
Qβ0 = 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,
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,
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
Answer (Detailed Solution Below)
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
(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'' +
q(x) =
now,
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
(b) has a solution sin(x) which has zeros at points x = 0, kπ, for k ∈ ℕ
(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
Which of the following statements is true?
Answer (Detailed Solution Below)
It has infinitely many solutions which are continuously differentiable on (0, ∞).
Initial Value Problem Question 6 Detailed Solution
Concept:
If
Explanation:
Here
∵ 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
Answer (Detailed Solution Below)
Initial Value Problem Question 7 Detailed Solution
Concept:
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
has infinitely many solution if
Answer (Detailed Solution Below)
Initial Value Problem Question 8 Detailed Solution
Concept:
If
Explanation:
Here
0 <
⇒ 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
Answer (Detailed Solution Below)
Initial Value Problem Question 9 Detailed Solution
Explanation:
which is linear in x
IF =
So, solution is
x
x =
x(0) = 1 ⇒ c = 1
hence
x(t) =
Therefore
x(1) =
Option (2) is true.
Initial Value Problem Question 10:
Consider the initial value problem
Which of the following statements is true?
Answer (Detailed Solution Below)
Initial Value Problem Question 10 Detailed Solution
Calculation:
Integrating both sides
Since given x(1) = 1 ⇒ -1 = 1 + c ⇒ c = -2
So
⇒ x =
As, x(t) → ∞ at t = 2
Therefore, solution blows up in a finite time.
The correct answer is option 1.