Question

The set of linearly independent solutions of the differential equation $(D^4 - D^3)y = 0; D = \frac{d}{dx}$ is

• A. $\{1, x, e^{-x}, x e^{-x}\}$
• B. $\{1, x, x^2, e^{-x}\}$
• C. $\{1, x, e^x, x e^x\}$
• D. $\{1, x, x^2, e^x\}$

Hint:

For repeated root $m = 0$ with multiplicity 3, the solutions are polynomials $1, x, x^2$ up to degree $3-1 = 2$. Combined with $e^x$, we get $\{1, x, x^2, e^x\}$.

Answer 1

The Correct Option is D

Step 1: Concept
For an $n$-th order linear homogeneous differential equation with constant coefficients $P(D)y = 0$, the set of $n$ linearly independent solutions corresponds to the fundamental basis generated by the roots of its auxiliary equation.

Step 2: Key Formulas and Approach
1. Write the auxiliary equation $P(m) = 0$. 2. If a real root $m = \alpha$ is repeated $k$ times, it generates $k$ linearly independent solutions: \[ e^{\alpha x}, \, x e^{\alpha x}, \, x^2 e^{\alpha x}, \, \dots, \, x^{k-1} e^{\alpha x} \]

Step 3: Step-by-step Explanation

• Given differential equation: \[ (D^4 - D^3)y = 0 \]
• Form the auxiliary equation: \[ m^4 - m^3 = 0 \]
• Factor the polynomial: \[ m^3 (m - 1) = 0 \]
• Find roots and their multiplicities: \[ m = 0 \text{ (repeated 3 times)}, \quad m = 1 \text{ (distinct root)} \]
• Construct linearly independent solutions for $m = 0$ ($k = 3$): \[ y_1 = e^{0x} = 1 \] \[ y_2 = x e^{0x} = x \] \[ y_3 = x^2 e^{0x} = x^2 \]
• Construct linearly independent solution for $m = 1$ ($k = 1$): \[ y_4 = e^{1x} = e^x \]
• The set of 4 linearly independent solutions is: \[ \{1, x, x^2, e^x\} \]

Step 4: Final Answer
The set of linearly independent solutions is $\{1, x, x^2, e^x\}$. Thus, Option (D) is correct.