Ordinary Differential Equations MCQ Quiz in తెలుగు - Objective Question with Answer for Ordinary Differential Equations - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Solution - The Green's Function for ODE 

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

G(x,t) =  

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

here the homogenous differential equation is 

y"+5y'+6y=0 

Now G(x,t) = 

 G(x,t) = e2(t-x) - e3(t-x)

Therefore, Correct Option is Option 2.

Solution:

 Given Ordinary Differential equation is 

Dividing (x-1) in L.H.S and R.H.S 

 = 0

Now for x = 0 

 (0/0 form)

Using L' Hospital we get 

 (0/0 form)

Again using L' Hospital form we get

L=  

now, 

for x=1 

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() =  sin(√6 ) = sin() = 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

 = f(x,y)

​ = 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

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) = 

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 =  and q = .

Calculation:

We have, 4xy" + 2y' + y = 0

⇒  

⇒ P(x) =  and Q(x) = 

⇒ x = 0 is a singular point.

Also,  =  =  = p

 =  = 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

⇒ 

⇒ 

⇒  [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, 

⇒ 

⇒ 

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

 = 0

⇒ 

 = 0

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

⇒ 

 = 0

In general, equating co-efficient of  to zero, we have

⇒ 

⇒ 

When m = 0:

, and so on.

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

⇒ 

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

∴ 

⇒ 

⇒ 

∴ 

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:

with boundary conditions 

Rewriting the equation:

The equation is simplified to

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

⇒ 

⇒   

CF = C₁e^z + C₂ze^z 

 i.e., CF = 

PI = 

   = 

  = 

  =  = 
 
+ 

Applying Boundary Conditions:

y(1) = 0:

 
 
⇒ 

: 

   
⇒  

⇒ 
   

⇒ 

⇒ 

Hence option 1) is correct.

Concept:

Picard’s Existence and Uniqueness Theorem: Consider the Initial Value Problem (IVP) , y(x₀) = y₀, suppose that f(x, y) and  are continuous functions in some open rectangle R = {(x, y): a 0, y₀) . Then the IVP has a unique solution in some closed interval I = [x₀ - h,x0 + h] where h > 0.

Explanation: 

 x ∈ ℝ, y(0) = y0,

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

(x, y) = - x sin(xy)

Both are continuous in a open rectangular region R = {(x, y): a 0) 

Now, |(x, y)| = |-x sin(xy)| = |x||sin(xy)| ≤ |x| |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. 

(∗) 0, \\ y(0)=y_{0}\end{array}\right.\)

(∗∗) 0, \\ z(0)=z_{0}\end{array}\right.\)

Let us assume α = 2

then (∗) ⇒

⇒  = dt

⇒  = t + c (integrating)

Using y(0) = y₀ we get

c = 

⇒  = t 

⇒ y = 

y is not defined if

1 - ty₀ = 0 ⇒ t =  > 0  as  y0 > 0

So (∗) does not have a global solution.

(1), (2) are false

(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