Question

The particular integral of $\left(D^4 - D^3 - 9D^2 - 11D - 4\right)y = e^{-x}$, where $D = \frac{d}{dx}$, is:}

• A. \(-\frac{x^2 e^{-x}}{20}\)
• B. \(-\frac{x e^{-x}}{15}\)
• C. \(-\frac{x^3 e^{-x}}{30}\)
• D. \(-\frac{e^{-x}}{10}\)

Hint:

Whenever $f(a)=0$, instead of factoring the entire polynomial, keep multiplying the expression by $x$ and differentiating the denominator with respect to $D$ until a non-zero denominator value is encountered upon substitution of $D=a$.

Answer 1

The Correct Option is C

Concept: The particular integral (P.I.) of a linear ordinary differential equation $f(D)y = e^{ax}$ is given by: \[ \text{P.I.} = \frac{1}{f(D)} e^{ax} \] If substituting $D = a$ yields $f(a) = 0$, this represents a case of failure. If $a$ is a root of multiplicity $r$ such that $f(D) = (D-a)^r \phi(D)$ where $\phi(a) \neq 0$, the general shortcut formula states: \[ \text{P.I.} = \frac{x^r}{r! \cdot \phi(a)} e^{ax} \] Alternatively, we can evaluate it step-by-step using successive shifts or successive differentiation of the denominator via the formula: $\frac{1}{f(D)}e^{ax} = x \frac{1}{f'(D)}e^{ax}$.

Step 1: Express the Particular Integral formula.
Here, $f(D) = D^4 - D^3 - 9D^2 - 11D - 4$ and the forcing function is $e^{-x}$, so $a = -1$. \[ \text{P.I.} = \frac{1}{D^4 - D^3 - 9D^2 - 11D - 4} e^{-x} \]

Step 2: Test for case of failure.
Substitute $D = -1$ into the denominator polynomial: \[ f(-1) = (-1)^4 - (-1)^3 - 9(-1)^2 - 11(-1) - 4 = 1 - (-1) - 9(1) + 11 - 4 \] \[ f(-1) = 1 + 1 - 9 + 11 - 4 = 0 \] Since the denominator is zero, this is a case of failure.

Step 3: Differentiate the operator using the rule $\text{P.I. = x \frac{1}{f'(D)} e^{ax}$.}
Let us find the first derivative of $f(D)$: \[ f'(D) = \frac{d}{dD}(D^4 - D^3 - 9D^2 - 11D - 4) = 4\text{D}^3 - 3\text{D}^2 - 18\text{D} - 11 \] Now, test substituting $D = -1$ into $f'(D)$: \[ f'(-1) = 4(-1)^3 - 3(-1)^2 - 18(-1) - 11 = -4 - 3 + 18 - 11 = 0 \] This is another case of failure.

Step 4: Differentiate again to find $f''(D)$.
Multiply by another $x$ and differentiate the denominator again: \[ \text{P.I.} = x^2 \frac{1}{f''(D)} e^{-x} \] \[ f''(D) = \frac{d}{dD}(4\text{D}^3 - 3\text{D}^2 - 18\text{D} - 11) = 12\text{D}^2 - 6\text{D} - 18 \] Now, test substituting $D = -1$ into $f''(D)$: \[ f''(-1) = 12(-1)^2 - 6(-1) - 18 = 12 + 6 - 18 = 0 \] Once again, it evaluates to zero.

Step 5: Differentiate a third time to find $f'''(D)$.
Multiply by another $x$ and differentiate the denominator again: \[ \text{P.I.} = x^3 \frac{1}{f'''(D)} e^{-x} \] \[ f'''(D) = \frac{d}{dD}(12\text{D}^2 - 6\text{D} - 18) = 24\text{D} - 6 \] Substitute $D = -1$ into $f'''(D)$: \[ f'''(-1) = 24(-1) - 6 = -24 - 6 = -30 \] Since this value is non-zero, the calculation stabilizes: \[ \text{P.I.} = x^3 \cdot \frac{1}{-30} e^{-x} = -\frac{x^3 e^{-x}}{30} \] This precisely corresponds to Option (C).