Solution of higher order homogeneous linear differential equations with constant coefficient

Correct Answer: a) y = C₁eˣ + C₂e²ˣ + C₃e³ˣ

Explanation: The characteristic equation for the given differential equation is:

$r^3-6r^2+11r-6=0$

Using factorization or any root-finding method, the roots are:

$r=1,r=2,r=3$

Since all roots are real and distinct, the general solution is:

$y=C_1{e}^x+C_2{e}^{\mathrm{2x}}+C_3{e}^{\mathrm{3x}}$

Therefore, the correct answer is: Option (a).

Correct Answer: b) Exponential and trigonometric terms

Explanation: The characteristic equation for the given differential equation is: \[ r^4 + 8r^2 + 16 = 0 \] Rearrange to find: \[ (r^2 + 4)^2 = 0 \] Taking the square root twice: \[ r = \pm 2i \text{ with multiplicity 2} \] Since the characteristic roots include both real and imaginary values, the general solution consists of both exponential and trigonometric terms: \[ y = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) + e^{\alpha x}(C_3 x \cos(\beta x) + C_4 x \sin(\beta x)) \] Therefore, the correct answer is: Option (b).

Correct Answer: a) r³ + 3r² - 4r - 12 = 0

Explanation: The characteristic equation of a differential equation is formed by substituting $y^{(n)}\Rightarrow r^n$ Using this approach, we get: $r^3+3r^2-4r-12=0$ Therefore, the correct answer is: Option (a).

Correct Answer: b) ±2i (each with multiplicity 2)

Explanation: The characteristic equation corresponding to the given differential equation is: \[ r^4 + 8r^2 + 16 = 0 \] Let \( u = r^2 \), then the quadratic becomes: \[ u^2 + 8u + 16 = 0 \] Which simplifies to: \[ (u + 4)^2 = 0 \] Factoring it, we get: \[ u = -4 \] Thus, \( r^2 = -4 \). Returning to \( r \) gives: \[ r = \pm 2i \] Since the root has multiplicity 2, the correct answer is: Option (b).

Correct Answer: b) y = (C₁ + C₂x + C₃x²)eˣ + C₄e⁻²ˣ

Explanation: The characteristic equation has roots \( r = 1 \) with multiplicity 3 and \( r = -2 \) with multiplicity 1. For a repeated root of multiplicity \( 3 \), the solutions take the form: \[ e^x, xe^x, x^2e^x \] And for the root \( r = -2 \), the corresponding solution is: \[ e^{-2x} \] Therefore, the general solution is: \[ y = (C_1 + C_2x + C_3x^2)e^x + C_4e^{-2x} \] Hence, the correct answer is: Option (b).

Correct Answer: b) y = (C₁ + C₂x + C₃x²)eˣ

Explanation: The characteristic equation of the given differential equation is: \[ r^3 - 3r^2 + 3r - 1 = 0 \] This is a cubic equation with a triple root at \( r = 1 \). For a triple root, the general form of the solution is: \[ y = (C_1 + C_2x + C_3x^2)e^x \] Therefore, the correct answer is: Option (b).

Correct Answer: b) y'''' - 2y''' + y'' = 0

Explanation: The given solution is: \[ y = C_1 + C_2x + (C_3 + C_4x)e^x \] - The presence of terms \( C_1 + C_2x \) suggests a double root at \( r = 0 \). - The terms \( (C_3 + C_4x)e^x \) indicate a repeated root at \( r = 1 \). Therefore, the characteristic equation must have roots 0 (with multiplicity 2) and 1 (with multiplicity 2), leading to the equation: \[ (r - 0)^2(r - 1)^2 = 0 \] Expanding this, we get: \[ r^4 - 2r^3 + r^2 = 0 \] Hence, the correct answer is: Option (b).

Correct Answer: d) 6e²ˣ

Explanation: The Wronskian \( W(e^x, e^{-x}, e^{2x}) \) is calculated using the determinant of the matrix formed by the functions and their derivatives:
\[ W = \begin{vmatrix} e^x & e^{-x} & e^{2x} \\ e^x & -e^{-x} & 2e^{2x} \\ e^x & e^{-x} & 4e^{2x} \end{vmatrix} \]
Evaluating this determinant, we get: \[ W = 6e^{2x} \] Therefore, the correct answer is: Option (d).

Correct Answer: c) 4

Explanation: The characteristic roots corresponding to the solutions are: \[ e^x \rightarrow r = 1, \quad xe^x \rightarrow r = 1 \text{ (multiplicity 2)}, \quad e^{-x} \rightarrow r = -1 \] The characteristic equation is formed as: \[ (r - 1)^2(r + 1) = 0 \] Expanding this: \[ (r^2 - 2r + 1)(r + 1) = r^3 - r^2 - 2r + 1 = 0 \] Since it is a third-degree polynomial, the minimal order of the ODE is 3.
Therefore, the correct answer is: Option (b).

Correct Answer: a) y = C₁ + C₂x + C₃e²ˣ + e³ˣ(C₄cosx + C₅sinx)

Explanation: The characteristic roots are given as 0 (with multiplicity 2), 2, and \( 3 \pm i \). The solution corresponding to the double root at 0 is: \[ C_1 + C_2x \] The solution for the root 2 is: \[ C_3e^{2x} \] For the complex roots \( 3 \pm i \), the general form of the solution is: \[ e^{3x}(C_4\cos{x} + C_5\sin{x}) \] Therefore, the general solution is: \[ y = C_1 + C_2x + C_3e^{2x} + e^{3x}(C_4\cos{x} + C_5\sin{x}) \]
Therefore, the correct answer is: Option (a).