Initial Value Problem MCQ Quiz in বাংলা - Objective Question with Answer for Initial Value Problem - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Existence of Solutions: The right-hand side \(\frac{\sin(y(x))}{1 + y(x)^4}. \) is smooth and well-behaved for all \(y \in \mathbb{R}\).

This ensures the existence of a global solution for any initial condition \( y_0 \in \mathbb{R}\) , and the solution exists for all \(x \in \mathbb{R}\) .

Explanation:

We are given the initial value problem (IVP)

\(y'(x) = \frac{\sin(y(x))}{1 + y(x)^4}, \quad x \in \mathbb{R}, \quad y(0) = y_0.\)

The right-hand side of the differential equation is:

\(\frac{\sin(y(x))}{1 + y(x)^4}. \)

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 \(\frac{\sin(y(x))}{1 + y(x)^4}. \) approaches 0 as |\(y(x)| \to \infty\) . 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 \(y_0\) . 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  \(y_0\) , 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 \(y_0\) .

There is no initial value \(y_0\) that leads to an unbounded solution.

Since the function on the right-hand side \(\frac{\sin(y(x))}{1 + y(x)^4}. \) is smooth and well-behaved for all \(y \in \mathbb{R}\) , there is a

unique solution to the IVP for any initial condition \(y_0 \in \mathbb{R}\) , and the solution exists for all \(x \in \mathbb{R}\) .

Option 1: There is a positive \(y_0\) such that the solution of the IVP is unbounded. False — all solutions are bounded, regardless of the initial value \(y_0\)
  
Option 2: There is a negative \(y_0\) such that the solution of the IVP is bounded. True — all solutions are bounded, including those with negative initial values.
  
Option 3: For every \(y_0 \in \mathbb{R}\) , every solution of the IVP is bounded. True — all solutions are bounded for any initial value \(y_0\)
  
Option 4: For every \(y_0 \in \mathbb{R}\) , there is a solution to the IVP for all \( x \in \mathbb{R}\) . 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).

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)

Explanation -

We have \(y^{\prime}=y+\frac{1}{2}|\sin \left(y^2)\right|,\), x > 0, y(0) = -1

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

Now \(y^{\prime}-y=\frac{1}{2}|\sin \left(y^2)\right| \le \frac{1}{2}\)

⇒ \(y^{\prime}-y= \frac{1}{2}\)

Now the solution of the differential equation is -

C.F. = C₁e^x

PI = \(\frac{1}{D-1} .\frac{1}{2} = - \frac{1}{2}\)

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 \(\displaystyle\lim _{x \rightarrow \alpha^{-}}|y(x)|=∞\) ⇒ 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.

Explanation:

the initial value problem

\(\frac{d y}{d x}=f(x, y)\), 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

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, \(P=\frac{3}{4}, Q=\frac{1}{4}\)

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, \(P=\frac{5}{6}, Q=\frac{1}{6}\)

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

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 \(\int_1^{∞}q(x)dx\) = ∞ 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'' + \(\frac{1+4x^2}{1+x^2}\)y = 0

q(x) = \(\frac{1+4x^2}{1+x^2}\)
now,

 \(\int_1^{∞}q(x)dx\\=\int_0^{\infty}\frac{1+4x^2}{1+x^2}dx\)

\(=\int _1^{\infty}(4-\frac{3}{1+x^2})dx\\=[4x-3\tan^{-1}x]_0^{\infty}=\infty\)

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 \(\frac{1+4x^2}{1+x^2}\) ≥ 1

(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  

Concept:

If \(\frac{dy}{dx}=y^n\), n ∈ (0, 1) and y(a) = 0, a ∈ \(\mathbb R\) then the differential equation has infinite number of independent solution.

Explanation:

Here \( \frac{d y}{d t}=y^{\frac{5 k}{5 k+2}}\), k ∈ \(\mathbb N\), y(0) = 0

∵ n =  \(\frac{5k}{5k+2}\) < 1 ∀ k ∈ \(\mathbb N\)

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

Option (3) is correct

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.

Concept:

If \(\frac{dy}{dx}=y^{α}\), α ∈ (0, 1) and y(a) = 0, a ∈ \(\mathbb R\) then the differential equation has infinite number of independent solution.

Explanation:

Here \( \frac{d y}{d t}=y^{\frac{2k}{k+2}}\), y(0) = 0 has infinitely many solution is

0 < \(\frac{2k}{k+2}\) < 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

Explanation:

\(\frac{d x}{d t}\) + (sin t) x = t, x(0) = 1

which is linear in x

IF  = \(e^{\int \sin t dt}=e^{-\cos t}\)

So, solution is

x\(e^{-\cos t}\) = ∫ t\(e^{-\cos t-t}\)dt + c

x = \(e^{\cos t}\)∫ t\(e^{-\cos t}\)dt + c\(e^{-\cos t}\)

x(0) = 1 ⇒ c = 1

hence

x(t) = \(e^{\cos t}\)\(\int te^{-\cos t}dt\) + \(e^{-\cos t}\)

Therefore

x(1) = \(e^{\cos 1}\) \(e^{-\cos 1}\) + \(e^{-\cos 1}\) = 1 + \(e^{-\cos 1}\)

Option (2) is true.

Calculation:

\(\rm \frac{dx}{dt}=x^2⇒ \frac{dx}{x^2}=dt\)

Integrating both sides

\(\rm -\frac{1}{x}=t+c\)

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

So

\(\rm -\frac{1}{x}=t-2\)

⇒ x = \(\frac1{2-t}\)

As, x(t) → ∞ at t = 2

Therefore, solution blows up in a finite time.

The correct answer is option 1.