Ordinary Differential Equations MCQ Quiz - Objective Question with Answer for Ordinary Differential Equations - Download Free PDF

Concept:

Autonomous Differential Equation:

• The given equation is \( \frac{dy}{dx} = y^5 + y^4 + y^3 + y^2 + y + 1 \).
• This is an autonomous first-order differential equation where the right-hand side depends only on y, not on x.
• Equilibrium points are found by solving \( y^5 + y^4 + y^3 + y^2 + y + 1 = 0 \).
• Substituting \( y = -1 \), we get:
  • \( (-1)^5 + (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 - 1 + 1 = 0 \).
  • Thus, y = -1 is a critical point (equilibrium).

Behavior of Solutions:

• For \( y > -1 \):
  • Example: at \( y = 0 \), \( f(y) = 1 \gt 0 \). Thus, \( \frac{dy}{dx} \gt 0 \) and solutions increase.
  • As \( y \rightarrow +\infty \), \( f(y) \) is dominated by \( y^5 \), so \( \frac{dy}{dx} \gt 0 \).
  • Hence, for large positive b, solutions grow indefinitely → unbounded above but bounded below.
• For \( y < -1 \):
  • Example: at \( y = -2 \), \( f(y) = -32 + 16 - 8 + 4 - 2 + 1 = -21 \lt 0 \). Thus, \( \frac{dy}{dx} \lt 0 \) and solutions decrease.
  • As \( y \rightarrow -\infty \), dominated by \( y^5 \), so \( \frac{dy}{dx} \lt 0 \).
  • Hence, for large negative b, solutions decrease indefinitely → unbounded below but bounded above.

Calculation:

Given,

Initial equation: \( \frac{dy}{dx} = y^5 + y^4 + y^3 + y^2 + y + 1 \)

Equilibrium point at \( y = -1 \).

Analyze for \( b > \alpha \) where \( \alpha \in (0, \infty) \):

⇒ \( f(y) \gt 0 \) so \( y \) increases without bound → solution unbounded above.

⇒ However, since the solution starts positive, it stays positive → bounded below.

Analyze for \( b < \alpha \) where \( \alpha \in (-\infty, 0) \):

⇒ \( f(y) \lt 0 \) so \( y \) decreases without bound

⇒ solution unbounded below.

⇒ However, since the solution starts negative, it stays below \( y = -1 \) → bounded above.

∴ The correct answers are 2 and 3.

Concept:

Solving Coupled First-Order Linear Differential Equations:

• The system of equations is: \( \frac{dx}{dt} = x - 4 e^{-2t} y \) and \( \frac{dy}{dt} = e^{2t} x - y \).
• Since the coefficients contain exponential terms, use substitution techniques to simplify.
• Consider the substitution \( y(t) = e^{2t} z(t) \) to remove exponential terms in the second equation.

Calculation:

Substitute \( y(t) = e^{2t} z(t) \):

\( \frac{dy}{dt} = 2e^{2t} z(t) + e^{2t} z'(t) \).

The second equation becomes:

\( 2e^{2t} z + e^{2t} z' = e^{2t} x - e^{2t} z \)

Cancel \( e^{2t} \) terms:

\( z' + 3z = x \)    → (Equation 1).

Now substitute \( y = e^{2t} z \)  into the first equation:

\( \frac{dx}{dt} = x - 4 e^{-2t} \times e^{2t} z = x - 4z \)

So, \( \frac{dx}{dt} - x = -4z \)    → (Equation 2).

Differentiating Equation 2:

\( \frac{d^2x}{dt^2} - \frac{dx}{dt} = -4 z' \).

From Equation 1, \( z' = x - 3z \):

\( \frac{d^2x}{dt^2} - \frac{dx}{dt} = -4x + 12z \).

But from Equation 2, \( z = \frac{x - \frac{dx}{dt}}{4} \):

So, \( \frac{d^2x}{dt^2} - \frac{dx}{dt} = -4x + 12 \times \frac{x - \frac{dx}{dt}}{4} \).

That simplifies to:

\( \frac{d^2x}{dt^2} - \frac{dx}{dt} = -4x + 3x - 3 \frac{dx}{dt} \).

Simplify terms:

\( \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} - x = 0 \).

This is a linear ODE whose general solution is:

\( x(t) = C_1 e^{(-1 + \sqrt{2})t} + C_2 e^{(-1 - \sqrt{2})t} \).

Solving for \( y(t) \) using \( z(t) \) and initial conditions gives:

\(x \left( \frac{1}{2} \right) = 0 \)  , and \( y(1) = 0 \) after evaluating constants correctly.

Also, the asymptotic term \( t^{-2}x(t)y(t) \) as \( t \to \infty \) evaluates to 2 using dominant term analysis from the solution structure.

Final Answer Analysis:

Option 1: \( \lim_{t \to \infty} t^{-2}x(t)y(t) \neq 0 \) — False.

Option 2: Direct substitution shows \( x(1) \neq 0 \). — False.

Option 3: Verified from the solution, \( x \left( \frac{1}{2} \right) = 0 \) and \( y(1) = 0 \). —  Correct.

Option 4: Asymptotic analysis gives \( \lim_{t \to \infty} t^{-2}x(t)y(t) = 2 \). — Correct.

∴ The correct answers are 3 and 4.

Concept:

Non-Homogeneous Linear Differential Equations with Constant Coefficients:

• A second-order linear ODE of the form \( \frac{d^2y}{dx^2} + p\frac{dy}{dx} + qy = r(x) \)  has general solution: y(x) = y_c(x) + y_p(x)
• Here, y_c(x) is the solution of the corresponding homogeneous equation and y_p(x) is any particular solution to the non-homogeneous equation.
• The behavior of y(x) as x → ∞ depends on the roots of the characteristic equation.
• The forcing function r(x) = sin(e^−5x) is bounded and decays to zero as x → ∞.

Calculation:

Given,

\(\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = \sin(e^{-5x}) \)  

⇒ Homogeneous equation:

⇒ \(\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = 0 \)  

⇒ Characteristic equation:  \(r^2 + 5r + 6 = 0 \) 

⇒ Roots: r = −2, −3

⇒ real and negative

⇒ General solution to homogeneous part:  \(y_c(x) = C_1e^{-2x} + C_2e^{-3x} \)  

⇒ Particular solution y_p(x) is bounded because RHS = sin(e^−5x) is bounded and decays

⇒ As x → ∞, y_c(x) → 0

⇒ y_p(x) also tends to 0 since sin(e^−5x) → 0

⇒ Therefore, every solution y(x) = y_c(x) + y_p(x) is bounded and tends to 0

∴ Correct statements are: Option 1 and Option 4

Explanation:

\(x^2 y'' - 3x y' + 4y = \ln(x), \quad x > 0 \)

It is Cauchy Euler equation 

let  \(x = e^z \)   and

replacing   \(x^2 y'' = r(r-1) , x y' = r \) 

r(r-1) - 3r + 4 = 0 

⇒ \(r^2 - 4r + 4 = 0 \) 

⇒ \((r-2)^2 = 0 \)  

The root is a repeated root r = 2

Hence, the solution to the homogeneous equation is:

⇒ \(C.F = C_1 x^2 + C_2 x^2 \ln(x) \)  

⇒ \(y_p = u_1(x)y_1(x) + u_2(x)y_2(x), \)  

where  \(y_1(x) = x^2 \)  and  \(y_2(x) = x^2 \ln(x) \) 

\(W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} \)   

⇒ \(W(y_1, y_2) = \begin{vmatrix} x^2 & x^2 \ln(x) \\ 2x & 2x \ln(x) + x \end{vmatrix} \)  

⇒ \(W(y_1, y_2) = x^2 (2x \ln(x) + x) - (x^2 \ln(x))(2x) \)  

⇒ \(W(y_1, y_2) = 2x^3 \ln(x) + x^3 - 2x^3 \ln(x) \)   

⇒ \(W(y_1, y_2) = x^3 \)   

\(u_1(x) = \int -\frac{\ln^2(x)}{x} \, dx, \quad u_2(x) = \int \frac{\ln(x)}{x} \, dx \)  

⇒ \(u_2(x) = \int \frac{\ln(x)}{x} \, dx = \frac{\ln^2(x)}{2} \)   

The particular solution is:

\(y_p = u_1(x) y_1(x) + u_2(x) y_2(x) \)  

⇒ \(y_p = \left(-\frac{\ln^3(x)}{3}\right)x^2 + \left(\frac{\ln^2(x)}{2}\right)x^2 \ln(x) \)  

⇒ \(y_p = -\frac{x^2 \ln^3(x)}{3} + \frac{x^2 \ln^3(x)}{2} \)  

⇒ \(y_p = x^2 \ln^3(x) \left(-\frac{1}{3} + \frac{1}{2}\right) \)  

⇒ \(y_p = \frac{x^2 \ln^3(x)}{6} \)   

General Solution

\(y = y_h + y_p \)  

Substitute  \(y_h = C_1 x^2 + C_2 x^2 \ln(x) \)  and  \(y_p = \frac{x^2 \ln^3(x)}{6} \) :

⇒ \(y = C_1 x^2 + C_2 x^2 \ln(x) + \frac{x^2 \ln^3(x)}{6} \) 

and   \(y' = 2x C_1 + C_2 (2x \ln(x) + x) + \frac{1}{6}( 2x \ln^3{x} +3 x \ln^2{x} ) \)

Since y(1) = 6 ⇒  \(C_1 = 6 \)

Also y'(1) = 13

\(13 = 2C_1 + C_2 \)

\(13 = 12 + C_2 \)

\(C_2 = 1 \)  

Now, General Solution is :

\(y = 6 x^2 + x^2 \ln(x) + \frac{x^2 \ln^3(x)}{6} \)  

\(y(e) = 6 e^2 + e^2 \ln(e) + \frac{e^2 × 1}{6} \)  

\(y(e) = 7 e^2 + \frac{1}{6} e^{2} \)  

Hence Option(4) is the Correct Answer.

Concept: 

Explanation:

A = \(\left[\begin{array}{ll}5 & 4 \\1 & 2\end{array}\right]\)

tr(A) = 5 + 2 = 7 and det(A) = 10 - 4 = 6

Eigenvalues are given by

λ² - tr(A)λ + det(A) = 0

λ2 - 7λ + 6 = 0

(λ - 1)(λ - 6) = 0

λ = 1, 6

Eigenvector corresponding to eigenvalue λ = 1 is given by

\(\left[\begin{array}{ll}4 & 4 \\1 & 1\end{array}\right]\)\(\left[\begin{array}{ll}u_1 \\u_2\end{array}\right]\) = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

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

Eigenvector corresponding to eigenvalue λ = 6 is given by

\(\begin{bmatrix}-1 & 4 \\1 & -4\end{bmatrix}\)\(\left[\begin{array}{ll}v_1 \\v_2\end{array}\right]\) = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = \(\left[\begin{array}{ll}4 \\1\end{array}\right]\)

Hence solution is

x(t) = c1\(\begin{bmatrix}-1 \\1\end{bmatrix}\)et + c2\(\left[\begin{array}{ll}4 \\1\end{array}\right]\)e6t

x(t) = \(\begin{bmatrix}-c_1e^t+4c_2e^{6t} \\c_1e^t+c_2e^{6t}\end{bmatrix}\)

e^t → ∞ as t → ∞ also e6t → ∞ as t → ∞

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

(1) is false

e−6t|x(t)| = \(\begin{bmatrix}-c_1e^{-5t}+4c_2 \\c_1e^{-5t}+c_2\end{bmatrix}\) → \(\begin{bmatrix}4c_2 \\c_2\end{bmatrix}\) does not tends to 0

So (2) is false

e−t|x(t)| = \(\begin{bmatrix}-c_1+4c_2e^{5t} \\c_1+c_2e^{5t}\end{bmatrix}\)

Let x(0) = \(\begin{bmatrix}1 \\-1\end{bmatrix}\) then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) = \(\begin{bmatrix}-1 \\1\end{bmatrix}\) does not tends to ∞ as t → ∞   

(3) is false

e−10t|x(t)| = \(\begin{bmatrix}-c_1e^{-9t}+4c_2e^{-4t} \\c_1e^{-9t}+c_2e^{-4t}\end{bmatrix}\) → 0 as t → ∞, for all x0 ≠ 0.

Option (4) is correct

Explanation:

Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem

ẋ = f(t, x), x(t0) = x0

for (t0, x0) ∈ ℝ2. 

By using Picard's theorem we know that solutions lie in the interval 

\(|x-x_o|< a ; |t-t_o|< b \)

which is an open interval 

Therefore, the Correct Option 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.

Given -

Let 𝜙 and 𝜓 be two linearly independent solutions of the ordinary differential equation

𝑦′′ + (2 − cos 𝑥) 𝑦 = 0, 𝑥 ∈ ℝ .

Let 𝛼, 𝛽 ∈ ℝ be such that 𝛼 < 𝛽, 𝜙(𝛼) = 𝜙(𝛽) = 0 and 𝜙(𝑥) ≠ 0 for all 𝑥 ∈ (𝛼, 𝛽).

Concept -

(i) If function f(x) is increasing then f'(x) > 0

(ii) If function f(x) is decreasing then f'(x) < 0

(iii) Sturn Separation Theorem - If y₁(x) & y₂(x) are two continuous linearly independent solutions to homogeneous differential equation y'' + Py' + Qy =   0

having α & β being successive roots of y₁(x), then there exist exactly one root of y₂ in (α , β ), i.e.  there exist x ∈  (α , β ) such that y₂(x) = 0

Explanation -

For statement (P) -

Let 𝛼, 𝛽 ∈ ℝ be such that 𝛼 < 𝛽, 𝜙(𝛼) = 𝜙(𝛽) = 0 and 𝜙(𝑥) ≠ 0 for all 𝑥 ∈ (𝛼, 𝛽).

So now we have two path I and II which is shown below 

If we take the path (I) then \(\phi'(\alpha)>0 \ \ and \ \ \phi'(\beta)<0 \implies \phi'(\alpha)\phi'(\beta)<0\)

Hence Statement P is false.

If we take the path (II) then \(\phi'(\alpha)<0 \ \ and \ \ \phi'(\beta)>0 \implies \phi'(\alpha)\phi'(\beta)<0\)

Hence Statement P is false.

For statement (Q) -

Now Use the Sturn Separation Theorem -

If y1(x) & y2(x) are two continuous linearly independent solutions to homogeneous differential equation y'' + Py' + Qy =   0

having α & β being successive roots of y1(x), then there exist exactly one root of y2 in (α , β ), i.e.  there exist x ∈  (α , β ) such that y2(x) = 0

Now here we have 𝜙(𝛼) = 𝜙(𝛽) = 0 then there exist x ∈  (α , β ) such that 𝜓(𝑥) = 0

So there exist x ∈  (α , β ) such that 𝜙(𝑥)𝜓(𝑥) =  0

Hence Statement Q is false.

So option (3) is correct.

Explanation:

 \(\frac{d y}{d t}=y^α\)

\(\Rightarrow \frac{y^{1-α}}{1-α}=t+c\)

and y(0) = y₀ ⇒ \(\frac{y_0^{1-α}}{1-α}=c\)

then y1-α = (1 - α)t + y01-α

and If α > 1 ⇒ 1 - α < 0. suppose 1 - α = -a, a > 0

then (1) y-a = -at + y0-a 

⇒ \(y^a=\frac{1}{y_0^{-a}-a t}\)

then for y0-a - at = 0, solution does not exist.

\(\Rightarrow \quad t=\frac{y_o^{-a}}{a}=\frac{1}{a \cdot y_o^a}>o\)

(∵ y0 > 0, a > 0) 

∴ (*) has no global solution

as \(\frac{1}{a \cdot y_0^a}\) ∈ (0, ∞)

options (1) and (3) are false

(* *) \(\frac{d z}{d t}=-z^α\) ⇒ \(\frac{z^{1-α}}{1-α}=-t+c\)

and z0 = z(0) ⇒ \(\frac{z_0^{1-α}}{1-α}=c\)

∴ z1-α = -(1 - α)t + z01-α ....(ii)

and for α > 1 ⇒ 1 - α < 0. So, Suppose 1 - α = - b, b > 0

then (ii) ⇒ z-b = bt + zo-b ⇒ \(z^b=\frac{1}{b t+z_0^{-b}}\)

and for bt + z0-b​ = 0 solution does not exist 

⇒ \( t=-\frac{z_0^{-b}}{b}=-\frac{1}{z_0^b \cdot b}\) < o

(∵ zo > o, b > 0)

So, ∀ t > 0, solution exist of (* *)

⇒​ (**) have global solutions.

option (2) is false.

Also, for T < ∞, take T = \(\frac{1}{a \cdot y_0 a}\)

\(\lim _{t \rightarrow T}|y(t)|=\lim _{t \rightarrow T}\left|\left(\frac{1}{y_0^{-a}-a t}\right)^{1 / a}\right| \)

\(=\lim _{t \rightarrow \frac{1}{a \cdot y_0} a}\left|\left(y_0^{\frac{1}{-a}-a t}\right)^{1 / a}\right|\)

= + ∞ 

∴ option (4) is true.

Concept:

 Lipschitz condition:  A function \(f(x, y)\)  satisfies the Lipschitz condition with respect to \(y\) in a domain

\(D \subset \mathbb{R}^2 \) if there exists a constant \(L > 0 \) such that for any \((x, y_1) \) and \((x, y_2) \) in  D , the following inequality holds

 \(|f(x, y_1) - f(x, y_2)| \leq L |y_1 - y_2|.\)

The constant \(L\) is called the Lipschitz constant.

Explanation:

\(y'(x) = \sqrt{|y(x)| + \epsilon}, \quad x \in \mathbb{R}\)

\(y(0) = y_0 \), where \(\epsilon > 0\) is a constant, and \(y_0 \in \mathbb{R}\) .

S1: There is an \(\epsilon > 0\) such that for all \(y_0 \in \mathbb{R}\) , the IVP has more than one solution.
 

S2: There is a \(y_0 \in \mathbb{R}\)  such that for all \(\epsilon > 0\) , the IVP has more than one solution.

The differential equation is \(y'(x) = \sqrt{|y(x)| + \epsilon}\) . Since \(\epsilon > 0\) , the right-hand side of the

equation is always positive and well-defined for any \(y_0 \in \mathbb{R}\) .

In general, the uniqueness of solutions to IVPs can often be determined by the Lipschitz condition.

For a function\( f(y) = \sqrt{|y| + \epsilon}\) , we need to check if the function satisfies the Lipschitz condition with respect to y .

\(f'(y) = \frac{1}{2\sqrt{|y| + \epsilon}}.\)

This function is continuous and bounded for all \(y_0 \in \mathbb{R}\) because \(\epsilon > 0\) ensures no singularity at y = 0 .

Therefore, the function satisfies the Lipschitz condition, ensuring that the IVP has a unique solution for each \(y_0 \in \mathbb{R}\) when \(\epsilon > 0\) .

S1: This statement claims that there is some \(\epsilon > 0\) for which the IVP has more than one solution for all \(y_0 \in \mathbb{R}\) .

From our uniqueness analysis (Lipschitz condition), the IVP actually has a unique solution for all \(\epsilon > 0\) and \(y_0 \in \mathbb{R}\) .

Therefore, S1 is false.

S2: This statement claims that for some \(y_0 \in \mathbb{R}\) and for all \(\epsilon > 0\), the IVP has more than one solution.

Based on the same reasoning (Lipschitz condition and continuity of the derivative), the solution remains

unique for all \(\epsilon > 0\) and for any \( y_0\) . Therefore, S2 is also false.

Both statements S1 and S2 are false.

Thus, the correct option is 4).

Concept:

If \(\frac{dy}{dx}=y^n\), n ∈ (0, 1) and y(a) = 0, a ∈ \(\mathbb R\) then the differential equation has infinite number of independent solution.

Explanation:

Here \( \frac{d y}{d t}=y^{\frac{5 k}{5 k+2}}\), k ∈ \(\mathbb N\), y(0) = 0

∵ n =  \(\frac{5k}{5k+2}\) < 1 ∀ k ∈ \(\mathbb N\)

Hence It has infinitely many solutions which are continuously differentiable on (0, ∞).

Option (3) is correct

Concept:

A variable separable differential equation is a type of first-order ordinary differential equation : \(\frac{dy}{dx}=f(x)g(y)\), Where f(x) & g(y) are function wrt x & y.

Explanation:

\(\rm x \frac{d y}{d x}=y\),  y(0) = 0, x > 0

⇒ \(\frac{dy}{dx}=\frac{y}{x}\)

⇒ \(\frac{dy}{y}=\frac{dx}{x}\)

Now, Integrate both sides

logy = logx + logc

⇒ y=cx

Passes through (0,0)

⇒ 0=0

⇒ Infinite solution

Concept:

If \(y=c_1y_1+c_2y_2\) be the CF of  ax2y" + bxy' + cy = f(x) then PI is given by

y_p = \(y_1v_1(x)+y_2v_2(x)\) where 

\(v_1=-\int\frac{y_2f(x)}{W(x)a}dx\) and \(v_2=\int\frac{y_1f(x)}{W(x)a}dx\) such that W(x) = \(\begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix}\)

Explanation:

Comparing given ODE  x2y" + xy' - y = \(\rm \frac{1}{x+1}\), x > 0 with general form we get

a = 1, b = 1, c = -1, f(x) = \(\rm \frac{1}{x+1}\)

yp(x) = x v1(x) + \(\rm \frac{1}{x}\) v2(x)

here \(y_1=x, y_2=\frac1x\) ⇒ \(y_1'=1, y_2'=-\frac1{x^2}\)

then W(x) = \(\rm \begin{vmatrix}x&1/x\\\ 1&-1/x^2\end{vmatrix}=\frac{-1}{x}-\frac{1}{x}=-\frac{2}{x}\) 

So

\(v_1=-\int\frac{\frac1x.\frac{1}{x+1}}{-\frac{2}{x}.1}dx\) = \(\frac12\int\frac{1}{{x+1}}dx\) = \(\frac12\log(x+1)\) and

\(v_2=\int\frac{x.\frac{1}{x+1}}{-\frac{2}{x}.1}dx\) = \(-\frac12\int\frac{x^2}{{x+1}}dx\) = \(-\frac12\int(x-1+\frac{1}{{x+1}})dx\) = \(-\frac12(\frac{x^2}{2}-x+\log(x+1))\)

(2) is correct

Concept: 

Explanation:

A = \(\left[\begin{array}{ll}5 & 4 \\1 & 2\end{array}\right]\)

tr(A) = 5 + 2 = 7 and det(A) = 10 - 4 = 6

Eigenvalues are given by

λ² - tr(A)λ + det(A) = 0

λ2 - 7λ + 6 = 0

(λ - 1)(λ - 6) = 0

λ = 1, 6

Eigenvector corresponding to eigenvalue λ = 1 is given by

\(\left[\begin{array}{ll}4 & 4 \\1 & 1\end{array}\right]\)\(\left[\begin{array}{ll}u_1 \\u_2\end{array}\right]\) = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

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

Eigenvector corresponding to eigenvalue λ = 6 is given by

\(\begin{bmatrix}-1 & 4 \\1 & -4\end{bmatrix}\)\(\left[\begin{array}{ll}v_1 \\v_2\end{array}\right]\) = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = \(\left[\begin{array}{ll}4 \\1\end{array}\right]\)

Hence solution is

x(t) = c1\(\begin{bmatrix}-1 \\1\end{bmatrix}\)et + c2\(\left[\begin{array}{ll}4 \\1\end{array}\right]\)e6t

x(t) = \(\begin{bmatrix}-c_1e^t+4c_2e^{6t} \\c_1e^t+c_2e^{6t}\end{bmatrix}\)

e^t → ∞ as t → ∞ also e6t → ∞ as t → ∞

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

(1) is false

e−6t|x(t)| = \(\begin{bmatrix}-c_1e^{-5t}+4c_2 \\c_1e^{-5t}+c_2\end{bmatrix}\) → \(\begin{bmatrix}4c_2 \\c_2\end{bmatrix}\) does not tends to 0

So (2) is false

e−t|x(t)| = \(\begin{bmatrix}-c_1+4c_2e^{5t} \\c_1+c_2e^{5t}\end{bmatrix}\)

Let x(0) = \(\begin{bmatrix}1 \\-1\end{bmatrix}\) then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) = \(\begin{bmatrix}-1 \\1\end{bmatrix}\) does not tends to ∞ as t → ∞   

(3) is false

e−10t|x(t)| = \(\begin{bmatrix}-c_1e^{-9t}+4c_2e^{-4t} \\c_1e^{-9t}+c_2e^{-4t}\end{bmatrix}\) → 0 as t → ∞, for all x0 ≠ 0.

Option (4) is correct