System of ODE MCQ Quiz in বাংলা - Objective Question with Answer for System of ODE - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept: 

Explanation:

A = \(\left[\begin{array}{ll}2 & 2 \\1 & 3\end{array}\right]\)

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

\(\left[\begin{array}{ll}1 & 2 \\1 & 2\end{array}\right]\)\(\left[\begin{array}{ll}u_1 \\u_2\end{array}\right]\) = 0  

u1 + 2u2 = 0 ⇒ u1 = - 2u2 

Eigenvector is u = \(\begin{bmatrix}-2 \\1\end{bmatrix}\)

Eigenvector corresponding to eigenvalue λ = 4 is given by

\(\begin{bmatrix}-2 & 2 \\1 & -1\end{bmatrix}\)\(\left[\begin{array}{ll}v_1 \\v_2\end{array}\right]\) = 0  

v1 - v2 = 0 ⇒ v1 = v2 

Eigenvector is v = \(\left[\begin{array}{ll}1 \\1\end{array}\right]\)

Hence solution is

x(t) = c1\(\begin{bmatrix}-2 \\1\end{bmatrix}\)et + c2\(\left[\begin{array}{ll}1 \\1\end{array}\right]\)e4t

x(t) = \(\begin{bmatrix}-2c_1e^{t}+c_2e^{4t} \\c_1e^t+c_2e^{4t}\end{bmatrix}\)

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)| = \(\begin{bmatrix}-2c_1e^{-3t}+c_2 \\c_1e^{-3t}+c_2\end{bmatrix}\) → \(\begin{bmatrix}c_2 \\c_2\end{bmatrix}\) does not tends to 0 as t → ∞, for all x0 ≠ 0.

So (1) is false

e−t|x(t)| = \(\begin{bmatrix}-2c_1+c_2e^{3t} \\c_1+c_2e^{-3t}\end{bmatrix}\)→ \(\begin{bmatrix}-c_1 \\c_1\end{bmatrix}\) does not tends to 0 as t → -∞   

(2) is false

e−5t|x(t)| = \(\begin{bmatrix}-2c_1e^{-4t}+c_2e^{-t} \\c_1e^{-4t}+c_2e^{-t}\end{bmatrix}\) → 0 as t → ∞, for all x0 ≠ 0.

Option (3) is correct

Explanation:

Y' = BY, Y(0) = \(\left[\begin{array}{r} 2 \\ -1 \end{array}\right]\rm B=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right]\) and Y = \(\left[\begin{array}{ll} y_1(x) \\ y_2(x) \end{array}\right]\)

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

The eigenvector for 1 is given by

\(\left[\begin{array}{rr} 0 & 2 \\ 0 & -2 \end{array}\right]\begin{bmatrix}a_1\\a_2\end{bmatrix}\) = 0

⇒ a2 = 0

eigenvector for 1 is u = \(\left[\begin{array}{r} 1 \\ 0 \end{array}\right]\)

The eigenvector for -1 is given by

Concept:

The solution of the system of ODE of the form \(\frac{dX}{dt}=AX\) such that A is a square matrix is of the form

X = c₁v₁\(e^{λ_1t}\) +  c2v2\(e^{λ_2t}\) + .... +  cnvn\(e^{λ_nt}\) 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 \(\frac{dX}{dt}=AX\) where A = \(\begin{bmatrix}1&2\\0&-1\end{bmatrix}\) and X = \(\begin{bmatrix}x_1\\x_2\end{bmatrix}\)

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

Eigenvector corresponding to the eigenvalue 1 is given by

\(\begin{bmatrix}0&2\\0&-2\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\) ⇒ v = 0 so eigenvector = \(\begin{bmatrix}1\\0\end{bmatrix}\)

Eigenvector corresponding to the eigenvalue -1 is given by

\(\begin{bmatrix}2&2\\0&0\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\) ⇒ u + v = 0 so eigenvector = \(\begin{bmatrix}1\\-1\end{bmatrix}\)

So Solution is

\(\begin{bmatrix}x_1\\x_2\end{bmatrix}=c_1 e^t\begin{bmatrix}1\\0\end{bmatrix}+c_2 e^{-t}\begin{bmatrix}1\\-1\end{bmatrix}\)

\(\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}c_1 e^t+c_2 e^{-t}\\-c_2 e^{-t}\end{bmatrix}\)

\(x_1(0)=2, x_2(0)=-1\) ⇒ \(c_1+c_2=2, c_2=1\Rightarrow c_1=1\)

So solution is \(\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}e^t+e^{-t}\\-e^{-t}\end{bmatrix}\)

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 \(\frac{dy}{dx}+ Py = Q(x)\)  then,

\(I.F = e^{\int p dx}\)   

and the solution is \(y\times I.F = \int Q(I.F) dx +c\)

Option 1) Let n = 1 then,

x'(t) = Ax(t) + \(e^{λ_1t} v_1\)  

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

then x' = x + \(e^t\) 

⇒ x' - x = \(e^t\) which is linear differential Equation then, 

I.F= \(e^{\int{-dt}}\) = \(e^{-t}\) 

solution is \(x e^ {-t} \)  = \(\int(e^t e^{-t})\)dt + c

x = \(te^{t}+ce^{t}\) 

∴ 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:

\(y'' + 4y' + 5y = 0, \quad y(0) = 0, \quad y'(0) = 1\),

The characteristic equation for the given ODE is:

\(r^2 + 4r + 5 = 0.\)

Using the quadratic formula:

\(r = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{-4}}{2}\).

Simplify:

\(r = -2 \pm i.\)

Thus, the general solution of the differential equation is:

\(y(t) = e^{-2t} \left(C_1 \cos(t) + C_2 \sin(t)\right)\).

Initial Conditions
 At t = 0 , y(0) = 0 :
   
\( y(0) = e^{0} \left(C_1 \cos(0) + C_2 \sin(0)\right) = C_1 = 0\).
   

   Thus, the solution simplifies to:
   
 \( y(t) = e^{-2t} C_2 \sin(t).\)
   

 At t = 0 , y'(0) = 1 :
   First, compute y'(t) :
   
\( y'(t) = \frac{d}{dt}\left(e^{-2t} C_2 \sin(t)\right).\)
   
   Using the product rule:
   
\( y'(t) = e^{-2t} C_2 \cos(t) - 2e^{-2t} C_2 \sin(t).\)
   

   At t = 0 :
   
\( y'(0) = e^{0} C_2 \cos(0) - 2e^{0} C_2 \sin(0) = C_2 \cdot 1 - 2C_2 \cdot 0 = C_2\).
   

   Thus, C₂ = 1 .

The solution is:

\(y(t) = e^{-2t} \sin(t)\).
Substitute \(t = \frac{\pi}{2}\) :

\(y\left(\frac{\pi}{2}\right) = e^{-2 \cdot \frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right)\).

Simplify:
\( y\left(\frac{\pi}{2}\right) = e^{-\pi} \cdot 1 = e^{-\pi}\).

Using \(e^{-\pi} \approx 0.0432\) :

\(y\left(\frac{\pi}{2}\right) \approx 0.043.\)

Final Answer: 0.043.

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

We will update the solution of the question later.