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

Concept:

Autonomous Differential Equation:

• The given equation is .
• 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 .
• Substituting , we get:
  • .
  • Thus, y = -1 is a critical point (equilibrium).

Behavior of Solutions:

• For -1 \):
  • Example: at , . Thus, and solutions increase.
  • As , is dominated by , so .
  • Hence, for large positive b, solutions grow indefinitely → unbounded above but bounded below.
• For :
  • Example: at , . Thus, and solutions decrease.
  • As , dominated by , so .
  • Hence, for large negative b, solutions decrease indefinitely → unbounded below but bounded above.

Calculation:

Given,

Initial equation:

Equilibrium point at .

Analyze for \alpha \) where :

⇒ so increases without bound → solution unbounded above.

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

Analyze for where :

⇒ so decreases without bound

⇒ solution unbounded below.

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

∴ The correct answers are 2 and 3.

Concept:

Solving Coupled First-Order Linear Differential Equations:

• The system of equations is: and .
• Since the coefficients contain exponential terms, use substitution techniques to simplify.
• Consider the substitution to remove exponential terms in the second equation.

Calculation:

Substitute :

.

The second equation becomes:

Cancel terms:

   → (Equation 1).

Now substitute   into the first equation:

So,    → (Equation 2).

Differentiating Equation 2:

.

From Equation 1, :

.

But from Equation 2, :

So, .

That simplifies to:

.

Simplify terms:

.

This is a linear ODE whose general solution is:

.

Solving for using and initial conditions gives:

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

Also, the asymptotic term as evaluates to 2 using dominant term analysis from the solution structure.

Final Answer Analysis:

Option 1: — False.

Option 2: Direct substitution shows . — False.

Option 3: Verified from the solution, and . —  Correct.

Option 4: Asymptotic analysis gives . — 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   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,

⇒ Homogeneous equation:

⇒   

⇒ Characteristic equation:   

⇒ Roots: r = −2, −3

⇒ real and negative

⇒ General solution to homogeneous part:    

⇒ 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

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Explanation:

0 \)

It is Cauchy Euler equation 

let     and

replacing    

r(r-1) - 3r + 4 = 0 

⇒  

⇒   

The root is a repeated root r = 2

Hence, the solution to the homogeneous equation is:

⇒   

⇒   

where    and   

⇒   

⇒   

⇒    

⇒    

⇒    

The particular solution is:

⇒   

⇒   

⇒   

⇒    

General Solution

Substitute    and   :

⇒  

and   

Since y(1) = 6 ⇒  

Also y'(1) = 13

Now, General Solution is :

Hence Option(4) is the Correct Answer.

Concept: Solution of ODE x'(t) = Ax is x(t) = c₁ue^λ₁t + 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 + 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

 = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

Eigenvector is u = 

Eigenvector corresponding to eigenvalue λ = 6 is given by

 = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = 

Hence solution is

x(t) = c1et + c2e6t

x(t) = 

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)| =  →  does not tends to 0

So (2) is false

e−t|x(t)| = 

Let x(0) =  then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) =  does not tends to ∞ as t → ∞   

(3) is false

e−10t|x(t)| =  → 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 

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:

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.

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

Concept -

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

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

(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

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

If we take the path (I) then 0 \ \ and \ \ \phi'(\beta)

Hence Statement P is false.

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

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:

and y(0) = y₀ ⇒ 

then y1-α = (1 - α)t + y01-α

and If α > 1 ⇒ 1 - α 0

then (1) y-a = -at + y0-a 

⇒ 

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

o\)

(∵ y0 > 0, a > 0) 

∴ (*) has no global solution

as  ∈ (0, ∞)

options (1) and (3) are false

(* *)  ⇒ 

and z0 = z(0) ⇒ 

∴ z1-α = -(1 - α)t + z01-α ....(ii)

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

then (ii) ⇒ z-b = bt + zo-b ⇒ 

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

⇒  

(∵ zo > o, b > 0)

So, ∀ t > 0, solution exist of (* *)

⇒​ (**) have global solutions.

option (2) is false.

Also, for T , take T = 

= + ∞ 

∴ option (4) is true.

Concept:

 Lipschitz condition:  A function   satisfies the Lipschitz condition with respect to  in a domain

 if there exists a constant 0 \) such that for any  and  in  D , the following inequality holds

The constant  is called the Lipschitz constant.

Explanation:



, where  0\) is a constant, and  .

S1: There is an 0\) such that for all  , the IVP has more than one solution.
 

S2: There is a   such that for all  0\) , the IVP has more than one solution.

The differential equation is  . Since  0\) , the right-hand side of the

equation is always positive and well-defined for any  .

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

For a function , we need to check if the function satisfies the Lipschitz condition with respect to y .

This function is continuous and bounded for all  because  0\) ensures no singularity at y = 0 .

Therefore, the function satisfies the Lipschitz condition, ensuring that the IVP has a unique solution for each  when  0\) .

S1: This statement claims that there is some  0\) for which the IVP has more than one solution for all  .

From our uniqueness analysis (Lipschitz condition), the IVP actually has a unique solution for all  0\) and  .

Therefore, S1 is false.

S2: This statement claims that for some  and for all 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  0\) and for any  . Therefore, S2 is also false.

Both statements S1 and S2 are false.

Thus, the correct option is 4).

Concept:

If , n ∈ (0, 1) and y(a) = 0, a ∈  then the differential equation has infinite number of independent solution.

Explanation:

Here , k ∈ , y(0) = 0

∵ 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 : , Where f(x) & g(y) are function wrt x & y.

Explanation:

,  y(0) = 0, x > 0

⇒ 

⇒ 

Now, Integrate both sides

logy = logx + logc

⇒ y=cx

Passes through (0,0)

⇒ 0=0

⇒ Infinite solution

Concept:

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

y_p =  where 

 and  such that W(x) = 

Explanation:

Comparing given ODE  x2y" + xy' - y = , x > 0 with general form we get

a = 1, b = 1, c = -1, f(x) = 

yp(x) = x v1(x) +  v2(x)

here  ⇒ 

then W(x) =  

So

 =  =  and

 =  =  = 

(2) is correct

Concept: Solution of ODE x'(t) = Ax is x(t) = c₁ue^λ₁t + 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 + 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

 = 0  

u₁ + u₂ = 0 ⇒ u1 = - u2 

Eigenvector is u = 

Eigenvector corresponding to eigenvalue λ = 6 is given by

 = 0  

v1 - 4v2 = 0 ⇒ v1 = 4v2 

Eigenvector is v = 

Hence solution is

x(t) = c1et + c2e6t

x(t) = 

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)| =  →  does not tends to 0

So (2) is false

e−t|x(t)| = 

Let x(0) =  then

-c₁ + 4c₂ = 1 and c1 + c2 = -1

Solving them we get c1 = -1, c2 = 0

Hence x(t) =  does not tends to ∞ as t → ∞   

(3) is false

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

Option (4) is correct