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

Last updated on Jul 7, 2025

Latest Ordinary Differential Equations MCQ Objective Questions

Ordinary Differential Equations Question 1:

For b ∈ ℝ, let yb = yb(x) be the unique solution of the initial value problem 

\(\rm \frac{dy}{dx}=y^5+y^4+y^3+y^2+y+1\), y(0) = b

defined on its maximal interval of existence Ib. Then which of the following statements are true? 

  1. There exists an a ∈ (0, ∞) such that for every b ∈ ℝ with b > a, the solution yb is bounded above on Ib.
  2. There exists an a ∈ (0, ∞) such that for every b ∈ ℝ with b > a, the solution yb is bounded below on Ib​.
  3. There exists an a ∈ (-∞, 0) such that for every b ∈ ℝ with b < a, the solution yb is bounded above on Ib​.
  4. There exists an a ∈ (-∞, 0) such that for every b ∈ ℝ with b < a, the solution yb is bounded below on Ib​.

Answer (Detailed Solution Below)

Option :

Ordinary Differential Equations Question 1 Detailed Solution

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.

Ordinary Differential Equations Question 2:

If x = x(t), y = y(t) is the solution of the initial value problem \(\rm \frac{dx}{dt}=x-4e^{-2t}y\)

\(\rm \frac{dy}{dt}=e^{2t}x-y\)

x(0) = 1, y(0) = 1

then which of the following statements are true? 

  1. \(\rm \lim_{t\rightarrow \infty}t^{-2}x(t)y(t) = 0\)
  2. x(1) = 0, y(\(\frac{1}{2}\)) = 0
  3. x(\(\frac{1}{2}\)) = 0, y (1) = 0
  4. \(\rm \lim_{t\rightarrow \infty}t^{-2}x(t)y(t) = 2\)

Answer (Detailed Solution Below)

Option :

Ordinary Differential Equations Question 2 Detailed Solution

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.

Ordinary Differential Equations Question 3:

Consider the non-homogeneous ordinary differential equation (ODE) 

\(\rm \frac{d^2y}{dx^2}+5\frac{dy}{dx}+6y=\sin (e^{-5x)}, x>0\)

Then which of the following statements are true? 

  1. Every solution of the ODE is bounded on (0. ∞)  
  2. There exists a solution of the ODE which is unbounded on (0. ∞)  
  3. Every solution of the ODE is unbounded on (0. ∞)  
  4. Every solution of the ODE tends to zero as x → 

Answer (Detailed Solution Below)

Option :

Ordinary Differential Equations Question 3 Detailed Solution

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) = yc(x) + yp(x)
  • Here, yc(x) is the solution of the corresponding homogeneous equation and yp(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 yp(x) is bounded because RHS = sin(e−5x) is bounded and decays

⇒ As x → ∞, yc(x) → 0

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

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

∴ Correct statements are: Option 1 and Option 4

Ordinary Differential Equations Question 4:

Let L[y] = \(x^{2} \frac{d^{2} y}{d x^{2}}+p x \frac{d y}{d x}+q y\), where p, q are real constants. Let y1(x) and y2(x) be two solutions of L[y] = 0, x > 0, that satisfy y1(x0) = 1, \(y_{1}^{\prime}\)(x0) = 0, y2(x0) = 0 and \(y_{2}^{\prime}\)(x0) = 1 for some x0 > 0. Then, 

  1. y1(x) is not a constant multiple of y2(x) 
  2. y1(x) is a constant multiple of y2(x) 
  3. 1, ln x are solutions of L[y] = 0 when p = 1, q = 0
  4. x, ln x are solutions of L[y] = 0 when p + q ≠ 0

Answer (Detailed Solution Below)

Option :

Ordinary Differential Equations Question 4 Detailed Solution

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Ordinary Differential Equations Question 5:

The Given Differential equation is \(x^2 y'' - 3x y' + 4y = \ln(x), \quad x > 0 \) with conditions  y(1) = 6 and y'(1) = 13  then the value of y(e) is ?

  1. \(y(e) = 6 e^2 - \frac{1}{7} e^{2} \)
  2. \(y(e) = 6 e^2 + \frac{1}{7} e^{-2} \)
  3. \(y(e) = 7 e^2 - \frac{1}{6} e^{-2} \)
  4. \(y(e) = 7 e^2 + \frac{1}{6} e^{2} \)

Answer (Detailed Solution Below)

Option 4 : \(y(e) = 7 e^2 + \frac{1}{6} e^{2} \)

Ordinary Differential Equations Question 5 Detailed Solution

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.

Top Ordinary Differential Equations MCQ Objective Questions

Suppose x(t) is the solution of the following initial value problem in ℝ2

ẋ = Ax, x(0) = x0, where A = \(\left[\begin{array}{ll}5 & 4 \\1 & 2\end{array}\right]\).

Which of the following statements is true?

  1. x(t) is a bounded solution for some x0 ≠ 0.
  2. e−6t|x(t)| → 0 as t → ∞, for all x0 ≠ 0.
  3. e−t|x(t)| → ∞ as t → ∞, for all x0 ≠ 0.
  4. e−10t|x(t)| → 0 as t → ∞, for all x0 ≠ 0.

Answer (Detailed Solution Below)

Option 4 : e−10t|x(t)| → 0 as t → ∞, for all x0 ≠ 0.

Ordinary Differential Equations Question 6 Detailed Solution

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

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

et → ∞ 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

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. Suppose that J(t0, x0) represents the maximal interval of existence for the initial value problem. Which of the following statements is true?

  1. J(t0, x0) = ℝ.
  2. J(t0, x0) is an open set.
  3. J(t0, x0) is a closed set.
  4. J(t0, x0) could be an empty set.

Answer (Detailed Solution Below)

Option 2 : J(t0, x0) is an open set.

Ordinary Differential Equations Question 7 Detailed Solution

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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).

Let ϕ denote the solution to the boundary value problem (BVP) 

\(\rm \left\{\begin{matrix}(xy')'-2y'+\frac{y}{x}=1,&1

Then the value of ϕ(e) is

  1. \(\rm -\frac{e}{2}\)
  2. \(\rm -\frac{e}{3}\)
  3. \(\rm \frac{e}{3}\)
  4. e

Answer (Detailed Solution Below)

Option 1 : \(\rm -\frac{e}{2}\)

Ordinary Differential Equations Question 8 Detailed Solution

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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 = ez then it becomes

⇒ {D'(D' - 1) - D' + 1}y = ez

⇒ (D'2 - 2D' + 1)y = ez

Auxiliary equation is

m2 - 2m + 1 = 0

⇒ \((m-1)^2 = 0\)

⇒ \(m = 1,1 \)  

CF = C1ez + C2zez 

 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.

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).

Consider the following statements:

𝑃: 𝜙′ (𝛼)𝜙′ (𝛽) > 0.

𝑄: 𝜙(𝑥)𝜓(𝑥) ≠ 0 for all 𝑥 ∈ (𝛼, 𝛽).

Then 

  1. 𝑃 is TRUE and 𝑄 is FALSE
  2. 𝑃 is FALSE and 𝑄 is TRUE
  3. both 𝑃 and 𝑄 are FALSE
  4. both 𝑃 and 𝑄 are TRUE

Answer (Detailed Solution Below)

Option 3 : both 𝑃 and 𝑄 are FALSE

Ordinary Differential Equations Question 9 Detailed Solution

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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 

qImage6581565d3f94480fb493e4ed

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.

Let y0 > 0, z0 > 0 and α > 1.

Consider the following two differential equations:
\(\begin{aligned} &(*)\left\{\begin{array}{l} \frac{d y}{d t}=y^\alpha \quad \text { for } t>0, \\ y(0)=y_0 \end{array}\right. \\ &(* *)\left\{\begin{array}{l} \frac{d z}{d t}=-z^\alpha \quad \text { for } t>0, \\ z(0)=z_0 \end{array}\right. \end{aligned}\)
We say that the solution to a differential equation exists globally if it exists for all t > 0.

Which of the following statements is true?

  1. Both (*) and (**) have global solutions

  2. None of (*) and (**) have global solutions
  3. There exists a global solution for (*) and there exists a T < ∞  such that \(\lim _{t \rightarrow T}|z(t)|=+∞\)
  4. There exists a global solution for (**) and there exists a T < ∞ such that \(\lim _{t \rightarrow T}|y(t)|=+∞\)

Answer (Detailed Solution Below)

Option 4 : There exists a global solution for (**) and there exists a T < ∞ such that \(\lim _{t \rightarrow T}|y(t)|=+∞\)

Ordinary Differential Equations Question 10 Detailed Solution

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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.

Consider the initial value problem (IVP) 

\(\rm \left\{\begin{matrix}y'(x)=\sqrt{|y(x)+ε|}, & x ∈ R\\\ y(0)=y_0\end{matrix}\right.\)

Consider the following statements: 

S1: There is an ε > 0 such that for all y0 ∈ ℝ, the IVP has more than one solution.

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

Then

  1. both S1 and S2 are true
  2. S1 is true but S2 is false
  3. S1 is false but S2 is true
  4. both S1 and S2​ are false

Answer (Detailed Solution Below)

Option 4 : both S1 and S2​ are false

Ordinary Differential Equations Question 11 Detailed Solution

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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).

Let k be a positive integer. Consider the differential equation

\(\left\{\begin{aligned} \frac{d y}{d t} &=y^{\frac{5 k}{5 k+2}} \text { for } t>0, \\ y(0) &=0 \end{aligned}\right.\)
Which of the following statements is true?

  1. It has a unique solution which is continuously differentiable on (0, ∞)
  2. It has at most two solutions which are continuously differentiable on (0, ∞).
  3. It has infinitely many solutions which are continuously differentiable on (0, ∞).

  4. It has no continuously differentiable solution on (0, ∞)

Answer (Detailed Solution Below)

Option 3 :

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

Ordinary Differential Equations Question 12 Detailed Solution

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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

Ordinary Differential Equations Question 13:

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

  1. there will be no solution
  2. Unique solution
  3. only and only two solutions
  4. infinite solution

Answer (Detailed Solution Below)

Option 4 : infinite solution

Ordinary Differential Equations Question 13 Detailed Solution

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

Ordinary Differential Equations Question 14:

The P.I of x2y" + xy' - y = \(\rm \frac{1}{x+1}\), x > 0 is yp(x) = x v1(x) + \(\rm \frac{1}{x}\) v2(x) then v1 and v2 are given by

  1. \(v_1(x)=\frac{x^2}2\)\(v_2(x)=-\frac12(\frac{x^2}{2}+x+\log(x+1))\)
  2. \(v_1(x)=\frac12\log(x+1)\)\(v_2(x)=-\frac12(\frac{x^2}{2}-x+\log(x+1))\)
  3. \(v_1(x)=-\frac12\log(x+1)\), \(v_2(x)=\frac12(\frac{x^2}{2}-x+\log(x+1))\)
  4. None of these

Answer (Detailed Solution Below)

Option 2 : \(v_1(x)=\frac12\log(x+1)\)\(v_2(x)=-\frac12(\frac{x^2}{2}-x+\log(x+1))\)

Ordinary Differential Equations Question 14 Detailed Solution

Concept:

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

yp\(y_1v_1(x)+y_2v_2(x)\) 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

 

Ordinary Differential Equations Question 15:

Suppose x(t) is the solution of the following initial value problem in ℝ2

ẋ = Ax, x(0) = x0, where A = \(\left[\begin{array}{ll}5 & 4 \\1 & 2\end{array}\right]\).

Which of the following statements is true?

  1. x(t) is a bounded solution for some x0 ≠ 0.
  2. e−6t|x(t)| → 0 as t → ∞, for all x0 ≠ 0.
  3. e−t|x(t)| → ∞ as t → ∞, for all x0 ≠ 0.
  4. e−10t|x(t)| → 0 as t → ∞, for all x0 ≠ 0.

Answer (Detailed Solution Below)

Option 4 : e−10t|x(t)| → 0 as t → ∞, for all x0 ≠ 0.

Ordinary Differential Equations Question 15 Detailed Solution

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

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

et → ∞ 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

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