Question

Let $u, v$ and $w$ be the non-zero solutions of the differential equation
\[ (D^3 - 6D^2 + 11D - 6)y = 0; \quad D = \frac{d}{dx} \] Then the Wronskian of $u, v$ and $w$ is

• A. $e^{6x}$
• B. $2e^{6x}$
• C. $4e^{6x}$
• D. $6e^{6x}$

Hint:

Shortcut using Abel's Identity: $W(x) = C e^{-\int (-6) dx} = C e^{6x}$. For distinct exponential solutions $e^{\lambda_1 x}, e^{\lambda_2 x}, e^{\lambda_3 x}$, the constant $C = \prod_{i < j} (\lambda_j - \lambda_i) = (2-1)(3-1)(3-2) = 2$.

Answer 1

The Correct Option is B

Step 1: Concept
The Wronskian $W(u, v, w)(x)$ of three differentiable functions $u, v, w$ is defined by the determinant: \[ W(x) = \begin{vmatrix} u & v & w u' & v' & w' u'' & v'' & w'' \end{vmatrix} \] By Abel's Identity for a third-order homogeneous differential equation $y''' + p_1(x) y'' + p_2(x) y' + p_3(x) y = 0$, the Wronskian satisfies: \[ W(x) = W(x_0) \exp\left( -\int_{x_0}^x p_1(t) \, dt \right) \]

Step 2: Key Formulas and Approach
We can solve for the fundamental basis of solutions $u, v, w$ using the auxiliary polynomial equation: \[ m^3 - 6m^2 + 11m - 6 = 0 \] Once the linearly independent solutions $u, v, w$ are determined, we calculate their Wronskian determinant directly.

Step 3: Step-by-step Explanation

• Factor the auxiliary polynomial: \[ m^3 - 6m^2 + 11m - 6 = (m - 1)(m - 2)(m - 3) = 0 \]
• The roots are distinct real numbers: $m_1 = 1, m_2 = 2, m_3 = 3$.
• The corresponding linearly independent basis solutions are: \[ u(x) = e^x, \quad v(x) = e^{2x}, \quad w(x) = e^{3x} \]
• Compute derivatives of each solution: \[ u' = e^x, \quad u'' = e^x \] \[ v' = 2e^{2x}, \quad v'' = 4e^{2x} \] \[ w' = 3e^{3x}, \quad w'' = 9e^{3x} \]
• Construct and evaluate the Wronskian determinant: \[ W(x) = \begin{vmatrix} e^x & e^{2x} & e^{3x} e^x & 2e^{2x} & 3e^{3x} e^x & 4e^{2x} & 9e^{3x} \end{vmatrix} \]
• Factor out $e^x, e^{2x}, e^{3x}$ from the columns: \[ W(x) = e^x \cdot e^{2x} \cdot e^{3x} \begin{vmatrix} 1 & 1 & 1 1 & 2 & 3 1 & 4 & 9 \end{vmatrix} = e^{6x} \begin{vmatrix} 1 & 1 & 1 1 & 2 & 3 1 & 4 & 9 \end{vmatrix} \]
• The remaining $3 \times 3$ determinant is a Vandermonde determinant for $(1, 2, 3)$: \[ \begin{vmatrix} 1 & 1 & 1 1 & 2 & 3 1 & 4 & 9 \end{vmatrix} = (2 - 1)(3 - 1)(3 - 2) = 1 \cdot 2 \cdot 1 = 2 \]
• Thus, $W(x) = 2 e^{6x}$.

Step 4: Final Answer
The Wronskian of $u, v$ and $w$ is $2 e^{6x}$. Thus, Option (B) is correct.