Ordinary Differential Equations MCQ Quiz in मल्याळम - Objective Question with Answer for Ordinary Differential Equations - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Solution - The Green's Function for ODE 

P(x)y" + Q(x)y'+ r(x)y = f(x) is given by 

G(x,t) = \(\frac{-1}{P(t)} \frac{y_1(x)y_2(t)- y_1(t)y_2(x)}{y_1(t)y_2'(t)-y_2(t)y_1'(t)}\) 

y1 and y2 are L.I solution of Homogenous Differential equation 

here the homogenous differential equation is 

y"+5y'+6y=0 

\(m^2+5m+6= 0; (m+2)(m+3)= 0\)

\(y(x)= c_1e^{-2x}+c_2e^{-3x}\)

Now G(x,t) = \(-1\cdot\frac{(e^{-2x}e^{-3t}- e^{-2t}e^{-3x})}{e^{-2t}(-3e^{-3t})-e^{-3t}(-2e^{-2t})}\)

 G(x,t) = e2(t-x) - e3(t-x)

Therefore, Correct Option is Option 2.

Solution:

 Given Ordinary Differential equation is 

\((x-1)y''+(cot\pi x)y'+ (cosec^2\pi x)y = 0\)

Dividing (x-1) in L.H.S and R.H.S 

\(y''+\frac{(cot\pi x)y'}{x-1}+\frac {(cosec^2\pi x)y}{x-1}\) = 0

Now for x = 0 

\(\lim_{x\rightarrow0}\frac{ (x-0)(cot\pi x )}{x-1} \) (0/0 form)

Using L' Hospital we get 

\(\lim_{x\rightarrow 0} \frac{xcos\pi x}{(x-1)(sin\pi x)} \) (0/0 form)

Again using L' Hospital form we get

L= \(\frac{-1}{\pi}\) 

now, \(lim_{x\rightarrow0} \frac{x^2cosec^2\pi x}{x-1} = \frac{2}{-2\pi^2}\)

for x=1 

\(lim_{x\rightarrow 1}\frac{x-1cot\pi x}{x-1} = \infty\)

so, 0 is regular point where 1 is irregular 

Therefore Option 1 is correct . 

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:

If the eigenvalues of the Jacobian matrix at the critical point are negative then that critical point is an asymptotically stable node

Explanation:

Given system of ordinary differential equations

\(\frac{dx}{dt}=x(3-2x-2y),\)

\(\frac{dy}{dt}=y(2-2x-y),\)

So F(x,y) = x(3 - 2x - 2y) = 3x - 2x² - 2xy and G(x,y) = y(2 - 2x - y) = 2y - 2xy - y². 

F_x = 3 - 4x - 2y, F_y = - 2x, G_x = - 2y, G_y = 2 - 2x - 2y

At (0, 2), Fx = - 3 - 4 = - 1, Fy = 0, Gx = - 4, Gy = 2 - 0 - 4 = - 2

Hence at (0,2) Jacobian is

J(0,2) = \(\begin{bmatrix}-1&0\\-4&-2\end{bmatrix}\)

This is an upper triangular matrix so eigenvalues are -1, -2.

Both eigenvalues are negative so (0,2) is a stable node.

Option (3) 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: 

Second-Order Differential Equations: Solve the homogeneous equation first, then find the particular solution.

Boundary Value Problems: Use boundary conditions to determine the constants in the general solution.

Explanation:
\((xy')' - 2y' + \frac{y}{x} = 1, \quad 1 < x < e^4\)

with boundary conditions \(y(1) = 0, \quad y(e^4) = 4e^4\)

Rewriting the equation:

The equation is simplified to

 \(x^2 y'' - xy' + y = x\)

Let x = e^z then it becomes

⇒ {D'(D' - 1) - D' + 1}y = e^z

⇒ (D'² - 2D' + 1)y = e^z

Auxiliary equation is

m² - 2m + 1 = 0

⇒ \((m-1)^2 = 0\)

⇒ \(m = 1,1 \)  

CF = C₁e^z + C₂ze^z 

 i.e., CF = \( C_1 x + C_2 x \ln(x)\)

PI = \({1\over (D'^2-2D'+1)}e^z\)

   = \(e^z{1\over ((D'+1)^2-2(D'+1)+1)}.1\)

  = \(e^z{1\over D'^2}.1\)

  = \({z^2\over 2}e^z\) = \(\frac {x(\ln x)^2}{2}\)
 
\( y(x) = C_1 x + C_2 x \ln(x) \)+ \(\frac {x(ln x)^2}{2}\)

Applying Boundary Conditions:

y(1) = 0:

 \(C_1 \cdot 1 + C_2 \cdot 1 \ln(1) + \frac{1^2}{2}.0 = 0\)
 
⇒ \(C_1 = 0\)

\(y(e^4) = 4e^4 \): 

\(\frac{1}{2} e^4 + C_2 e^4 \ln(e^4) + e^4\frac{(ln e^4)^2}{2} = 4e^4\)
   
⇒  \( C_2 e^4 \cdot 4 + 8{e^4} = 4e^4\)

⇒ \(c_2 = -1\)
   
\(y(x) = -x ln x + \)\(\frac {x(ln x)^2}{2}\)

⇒ \(y(e) = -e +\frac{e}{2}\)

⇒ \(y(e) = - \frac{e}{2}\)

Hence option 1) is correct.

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

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