Ordinary Differential Equations MCQ Quiz in मराठी - Objective Question with Answer for Ordinary Differential Equations - मोफत PDF डाउनलोड करा

Explanation:

y(x) = v sec(x) ⇒ y' = v' sec x + v sec x tan x

⇒ y'' = v'' sec x + v' sec x tan x + v' sec x tan x + v sec x tan² x + v sec³ x

Substituting these values in the given differential equation 

y'' - (2tan x)y' + 5y = 0

⇒ v'' sec x + v' sec x tan x + v' sec x tan x + v sec x tan2 x + v sec3 x - 2 v' sec x tan x - 2 v sec x tan² x + 5v sec(x) = 0

⇒ v'' sec x + v sec3 x - v sec x tan2 x + 5v sec(x) = 0  

⇒ v'' sec x + v sec x(sec2 x - tan2 x + 5) = 0

⇒ v'' sec x + 6v sec x = 0 (∵ sec2 x - tan2 x = 1)

⇒ v'' + 6v = 0

⇒ v = c₁ cos(√6 x) + c₂ sin(√6 x) ...(i)

Given y(0) = 0 and y'(0) = √6 ⇒ v(0) = 0 and v'(0) = √6

Substituting initial conditions 

 v(0) = 0  ⇒ c1 = 0

So v = c2 sin(√6 x)

v' = c2 √6 cos(√6 x)

v'(0) = √6 ⇒ c2 √6 = √6 ⇒ c2 = 1

Hence v = sin(√6 x)

∴ v(\(\frac{\pi}{6\sqrt6}\)) =  sin(√6 \(\frac{\pi}{6\sqrt6}\)) = sin(\(\frac{\pi}{6}\)) = 0.5

∴ Option (3) is correct

Concept:

Bendixon's Criterion: If f_x and g_y are continuous in a simply connected region ℝ² and fx + gx ≠ 0 then the system of differential equations

\(\frac{dx}{dt}\) = f(x,y)

​\(\frac{dy}{dt}\) = g(x,y)​

has no closed trajectories inside ℝ

Explanation:

Here f(x,y) = 4x3y2 - x5y4  g(x,y) = x4y5 + 2x2y3

fx = 12x2y2 - 5x4y4, g_y = 5x4y4 + 6x2y2

Both fx and gy are continuous and 

 fx + gx = 12x2y2 - 5x4y4 + 5x4y4 + 6x2y2 = 18x2y2  ≠ 0 in whole ℝ2 as it is zero on (0,0) only.

Hence by Bendixsion Criterion, there is no closed path in ℝ2

Option (4) is correct.

Concept:

• Ordinary Point: A point x = x₀ is called an ordinary point of differential equation y'' + P(x)y' + Q(x) = 0, if P(x) and Q(x) are both analytical at x = x₀.

• A singular point x = x0 is called regular singular point if both (x - x₀)P(x) and (x - x₀)²Q(x) are analytic at x = x₀. Otherwise it is called irregular singular point.

• The indicial equation in variable m for regular singular point x₀ is represented by m(m - 1) + pm + q = 0, where p =\(\lim_{x\to x_0}(x - x_0)P(x)\)  and q = \(\lim_{x\to x_0}(x - x_0)^2Q(x)\).

Calculation:

We have, 4xy" + 2y' + y = 0

⇒ \(y''+\frac{1}{2x}\frac{dy}{dx}+\frac{1}{4x}y=0\) 

⇒ P(x) = \(\frac{1}{2x}\) and Q(x) = \(\frac{1}{4x}\)

⇒ x = 0 is a singular point.

Also, \(\lim_{x\to 0}xP(x)\) = \(\lim_{x\to 0}x(\frac{1}{2x})\) = \(\frac{1}{2}\) = p

\(\lim_{x\to 0}x^2Q(x)\) = \(\lim_{x\to 0}x^2(\frac{1}{4x})\) = 0 = q

⇒ x = 0 is a regular singular point.

Now, indicial equation for the given differential equation is given by m(m - 1) + pm + q = 0

⇒ \(m^2-m+\frac{m}{2}=0\)

⇒ \(m^2-\frac{m}{2}=0\)

⇒ \(m = 0, \frac{1}{2}\) [Distinct roots]

Therefore, we get two independent solutions corresponding to two different value of m.

Since, x = x₀ is regular singular point, we have to use Forbenious method to get the required solution.

Let, \(y =\sum_{n=0}^{\infty} a_n x^{m+n}\)

⇒ \(y' =\sum_{n=0}^{\infty}(m+n) a_n x^{m+n-1}\)

⇒ \(y'' =\sum_{n=0}^{\infty}(m+n)(m+n-1) a_n x^{m+n-2}\)

Substituting the values of y, y' and y" in the given equation, we have,

\(4 x \sum_{n=0}^{\infty}(m+n)(m+n-1) a_n x^{m+n-2}\)

\(+2 \sum_{n=0}^{\infty}(m+n) a_n x^{m+n-1}+\sum_{n=0}^{\infty} a_n x^{m+n}\) = 0

⇒ \(\sum_{n=0}^{\infty} 4(m+n)(m+n-1) a_n x^{m+n-1}\)

\(+\sum_{n=0}^{\infty} 2(m+n) a_n x^{m+n-1}+\sum_{n=0}^{\infty} a_n x^{m+n}\) = 0

Shifting the index of first two terms to m+n, we have

⇒ \(\sum_{n=0}^{\infty} 4(m+n+1)(m+n) a_{n+1} x^{m+n}\)

\(+\sum_{n=0}^{\infty} 2(m+n+1) a_{n+1} x^{m+n} +\sum_{n=0}^{\infty} a_n x^{m+n}\) = 0

In general, equating co-efficient of \(x^{m+n}\) to zero, we have

⇒ \([4(m+n+1)(m+n)+2(m+n+1)] a_{n+1}+a_n=0\)

⇒ \(a_{n+1}=\frac{a_n}{[4(m+n+1)(m+n)+2(m+n+1)]}, n \geq 0\)

When m = 0:

\(a_{n+1}=\frac{a_n}{[4(n+1)(n)+2(n+1)]}, n\geq0\)

\(a_1=\frac{a_0}{2}\)

\(a_2=\frac{a_1}{12}=\frac{a_0}{24}\), and so on.

Therefore, when m=0, one of the solution of y(x) is

\(y(x)=x^0(a_0+a_1 x+a_2 x^2+\cdots )\)

⇒ \(y(x)=a_0+\frac{a_0}{2}+\frac{a_0}{24} x^2+\cdots .\)

Substituting the initial condition y(0) = 1, we get a₀ = 1

∴ \( y(x)=1+\frac{x}{2}+\frac{x^2}{24}+\cdots \)

⇒ \(y'(x)=\frac{1}{2}+\frac{x}{12}+\cdots\)

⇒ \(y''(x)=\frac{1}{12}+\underbrace{\cdots \cdots \cdots}_{\text {higher power of } x}\)

∴ \(y''(0)=\frac{1}{12}\)

The correct answer is Option 2.

Concept:

Picard’s Existence and Uniqueness Theorem: Consider the Initial Value Problem (IVP) \(\rm \frac{dy}{dx}=f(x, y)\), y(x₀) = y₀, suppose that f(x, y) and \(\frac{\partial f}{\partial y}\) are continuous functions in some open rectangle R = {(x, y): a < x < b, c < y < d} that contains the point (x₀, y₀) . Then the IVP has a unique solution in some closed interval I = [x₀ - h,x0 + h] where h > 0.

Explanation: 

\(\rm \frac{dy}{dx}=\cos(xy),\) x ∈ ℝ, y(0) = y0,

Here f(x, y) = cos(xy)

\(\frac{\partial f}{\partial y}\)(x, y) = - x sin(xy)

Both are continuous in a open rectangular region R = {(x, y): a < x < b, c < y < d} containing (0, y₀) 

Now, |\(\frac{\partial f}{\partial y}\)(x, y)| = |-x sin(xy)| = |x||sin(xy)| ≤ |x| < b (as |sin(xy)| ≤ 1 for all x, y ℝ)

Hence by Picard’s existence and uniqueness theorem, 

the given IVP has a unique solution

Option (1) is true

Solution: 

Given differential equation 

\(\frac{dy}{dx}+α y\) = 0, y(0) = 1 where \(α \in R\)

\(\frac{dy}{dx} = -α y \)

by using variable separable form 

\(\frac{dy}{y} = -α dx \)

Integrate, both sides 

log y = - αx + log c₁

y = \(c_1e^{-α x}\)  

Given initial condition y(1) = 0 then, \(c_1 = 1\) 

So, the solution is

y = \(e^{-α x}\)

(1): y(1) = \(e^{-α}\) ≠ 0 for any α

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

(1) is false

(2): \(\displaystyle\lim_{x \rightarrow \infty}\)y(x) = \(\displaystyle\lim_{x \rightarrow \infty}\)\(e^{-α x}\) = 0 for all α > 0

So, there is no unique α such that \(\displaystyle\lim_{x \rightarrow \infty}\) y(x) = 0

(2) is false

(3): y(2) = 1

⇒ \(e^{-2α}\) = 1 ⇒ α = 0 

There is a α such that y(2) = 1

(3) is false

(4): y(1) = 2

⇒ \(e^{-α}\) = 2 ⇒ - α = log(2) ⇒ α = - log(2) 

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

(4) is correct

Explanation:

y0 > 0, z0 > 0 and α > 1. 

(∗) \(\left\{\begin{array}{l}\frac{d y}{d t}=y^{α} \quad \text { for } t>0, \\ y(0)=y_{0}\end{array}\right.\)

(∗∗) \(\left\{\begin{array}{l}\frac{d z}{d t}=-z^{α} \quad \text { for } t>0, \\ z(0)=z_{0}\end{array}\right.\)

Let us assume α = 2

then (∗) ⇒

 \(\frac{d y}{d t}=y^{2}, y(0)=y_{0}\) 

⇒ \(\frac{dy}{y^2}\) = dt

⇒ \(-\frac{1}{y}\) = t + c (integrating)

Using y(0) = y₀ we get

c = \(-\frac{1}{y_0}\)

⇒ \(-\frac{1}{y}\) = t \(-\frac{1}{y_0}\)

⇒ y = \(-\frac{y_o}{1-ty_0}\)

y is not defined if

1 - ty₀ = 0 ⇒ t = \(\frac{1}{y_0}\) > 0  as  y0 > 0

So (∗) does not have a global solution.

(1), (2) are false

\(\lim_{t\to\frac{1}{y_0}}|y(t)|=+\infty\)

(4) is correct

If we check (∗∗) by taking α = 2 we can see that (3) is false

Concept:

(i) If y1 and y2 are linearly independent then W(x) ≠ 0 ∀ x

(ii) If y1 and y2 are linearly dependent then W(x) = 0 ∀ x

Explanation:

By direct result, (4) is correct only

Explanation:

(x(t)2 ≤ 2 + \(\int_{0}^{t} \) 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 + \(\int_{0}^{t} \) s ds

⇒ t2 ≤ 2 + t2/2

⇒ t2/2 < 2

⇒ t2 < 4

⇒ - 2 < t < 2

⇒ - 2 < x(t) < 2

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

(1) is correct 

Explanation:

\(\rm \left|\frac{dy}{dx}\right|+|y|=0\)

Let y(x) be any non-constant solution 

⇒ ∃ x ∈ \(\mathbb R\) such that

 y(x0) ≠ 0

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

then at x0, \(\rm \left|\frac{dy}{dx}\right|+|y|>0\) contradiction

So it has no non-constant solution

If y(x) = 0 then \(\rm \left|\frac{dy}{dx}\right|+|y|=0\)

So given ODE has only trivial solution 

(1) is correct

Concept:

Let f(x, y) be continuous in a domain D of the (x, y) plane and |f(x, y)| ≤ M such that M is constant. Let f(x, y) satisfy Lipschitz condition in y. Also let |x - x₀| ≤ h, |y - y₀| ≤ k lie in D, then the differential equation \(\frac{dy}{dx}=f(x,y)\) has unique solution y = y(x) for which y(x₀) = y₀

Explanation: 

Given differential equation can be written as

y' = \(\frac2t\)y + t²e^t

So f(t, y) =  \(\frac2t\)y + t2et

D = {(t, y) : 1 ≤ t ≤ 2, -∞ }

Now \(|\frac{\partial f}{\partial y}|=|\frac2t|\) ≤ 2 in D

So f(t, y) satisfies a Lipschitz condition in D. Hence given differential equation as a unique solution.

Option (1) is correct.