System of ODE MCQ Quiz in मराठी - Objective Question with Answer for System of ODE - मोफत PDF डाउनलोड करा

Concept: Solution of ODE x'(t) = Ax is x(t) = c1ueλ1t + c2veλ2t, where u, v are the eigenvectors corresponding to the eigenvalues λ1 and λ2 respectively and c1 and c2 are constants.

Explanation:

A = 

tr(A) = 5  and det(A) = 6 - 2 = 4

Eigenvalues are given by

λ2 - tr(A)λ + det(A) = 0

λ2 - 5λ + 4 = 0

(λ - 1)(λ - 4) = 0

λ = 1, 4

Eigenvector corresponding to eigenvalue λ = 1 is given by

 = 0  

u1 + 2u2 = 0 ⇒ u1 = - 2u2 

Eigenvector is u = 

Eigenvector corresponding to eigenvalue λ = 4 is given by

 = 0  

v1 - v2 = 0 ⇒ v1 = v2 

Eigenvector is v = 

Hence solution is

x(t) = c1et + c2e4t

x(t) = 

et → ∞ as t → ∞ also e4t → ∞ as t → ∞

So x(t) is not bounded solution for any x0 ≠ 0

(4) is false

e−4t|x(t)| =  →  does not tends to 0 as t → ∞, for all x0 ≠ 0.

So (1) is false

e−t|x(t)| = →  does not tends to 0 as t → -∞   

(2) is false

e−5t|x(t)| =  → 0 as t → ∞, for all x0 ≠ 0.

Option (3) is correct

Explanation:

Y' = BY, Y(0) =  and Y = 

B is an upper triangular matrix so the eigenvalues of B are 1, -1

The eigenvector for 1 is given by

 = 0

⇒ a2 = 0

eigenvector for 1 is u = 

The eigenvector for -1 is given by

Concept:

The solution of the system of ODE of the form  such that A is a square matrix is of the form

X = c₁v₁ +  c2v2 + .... +  cnvn where λ₁, λ2, ..., λn are the distinct real eigenvalues of A and v1, v2, ..., vn are corresponding eigenvectors.

Explanation:

Given system of differential equations can be written as  where A =  and X = 

Here A is an upper triangular matrix so eigenvalues are 1, -1

Eigenvector corresponding to the eigenvalue 1 is given by

 ⇒ v = 0 so eigenvector = 

Eigenvector corresponding to the eigenvalue -1 is given by

 ⇒ u + v = 0 so eigenvector = 

So Solution is

 ⇒ 

So solution is 

Option (1) is correct.

Explanation:

 Since f is locally Lipschitz function so f is bounded \Hence the ODE has unique solution in around a particular point.

So (2) false

Given (x1(0), x2(0)) = (1,1)

Hence there is a unique solution around time 0

Therefore (4) true, (1) true (3) false

Explanation:

When differential equations is in the form of   then,

and the solution is 

Option 1) Let n = 1 then,

x'(t) = Ax(t) +   

and Let A = I (Identity matrix) and λ₁ = 1, v₁ = 1 

then x' = x +  

⇒ x' - x =  which is linear differential Equation then, 

I.F=  =  

solution is   = dt + c

x =  

∴ Option 1 is correct.

By a similar argument in option 1, option 2 is incorrect.

Similarly, By the above explanation, Option 3 is correct.

By the above explanation, Option 4 is incorrect.  

The correct options are 1 and 3. 

Explanation:

The second order ordinary differential equation:

,

The characteristic equation for the given ODE is:

Using the quadratic formula:

.

Simplify:

Thus, the general solution of the differential equation is:

.

Initial Conditions
 At t = 0 , y(0) = 0 :
   
.
   

   Thus, the solution simplifies to:
   
 
   

 At t = 0 , y'(0) = 1 :
   First, compute y'(t) :
   

   
   Using the product rule:
   

   

   At t = 0 :
   
.
   

   Thus, C₂ = 1 .

The solution is:

.
Substitute :

.

Simplify:
.

Using :

Final Answer: 0.043.

The correct answers are (1), (2) & (4)

We will update the solution of the question later.