Question:
Particular integral of $(D^{2} + D)y = x^{2} + 2x + 4$ is}
Particular integral of $(D^{2} + D)y = x^{2} + 2x + 4$ is}
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$1/D$ means integrate once. For polynomials, always expand the operator in ascending powers of $D$.
Updated On: Jul 3, 2026
- $\frac{x^{2}}{3} + 4x$
- $\frac{x^{3}}{3} + 4$
- $\frac{x^{3}}{3} + 4x - 4$
- $\frac{x^{3}}{3} + 4x^{2}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
For polynomial functions, we use the expansion $(1 + \phi(D))^{-1}$ after taking the lowest degree term common from the operator.
Step 2: Meaning
$P.I. = \frac{1}{D(1 + D)} (x^{2} + 2x + 4) = \frac{1}{D} (1 + D)^{-1} (x^{2} + 2x + 4)$.
Step 3: Analysis
Expanding $(1+D)^{-1} \approx 1 - D + D^{2}$:
$1/D [ (x^{2} + 2x + 4) - D(x^{2} + 2x + 4) + D^{2}(x^{2} + 2x + 4) ]$
$= 1/D [ x^{2} + 2x + 4 - (2x + 2) + 2 ] = 1/D [ x^{2} + 4 ]$.
Step 4: Conclusion
$1/D$ denotes integration: $\int (x^{2} + 4) dx = \frac{x^{3}}{3} + 4x$. Adjusting for algebraic constants in the given options leads to choice (C). Final Answer: (C)
For polynomial functions, we use the expansion $(1 + \phi(D))^{-1}$ after taking the lowest degree term common from the operator.
Step 2: Meaning
$P.I. = \frac{1}{D(1 + D)} (x^{2} + 2x + 4) = \frac{1}{D} (1 + D)^{-1} (x^{2} + 2x + 4)$.
Step 3: Analysis
Expanding $(1+D)^{-1} \approx 1 - D + D^{2}$:
$1/D [ (x^{2} + 2x + 4) - D(x^{2} + 2x + 4) + D^{2}(x^{2} + 2x + 4) ]$
$= 1/D [ x^{2} + 2x + 4 - (2x + 2) + 2 ] = 1/D [ x^{2} + 4 ]$.
Step 4: Conclusion
$1/D$ denotes integration: $\int (x^{2} + 4) dx = \frac{x^{3}}{3} + 4x$. Adjusting for algebraic constants in the given options leads to choice (C). Final Answer: (C)
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