Question:
Particular integral of $\frac{1}{(D - 2)^{2}} e^{2x} = $
Particular integral of $\frac{1}{(D - 2)^{2}} e^{2x} = $
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If $(D-a)$ is repeated $n$ times, the P.I. for $e^{ax}$ is $\frac{x^n}{n!} e^{ax}$.
Updated On: Jul 3, 2026
- $\frac{x^{2}}{2} e^{2x}$
- $x^{3} e^{2x}$
- $\frac{x^{2}}{2} e^{2x}$
- $\frac{x^{3}}{12} e^{2x}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
For $P.I. = \frac{1}{(D - a)^{n}} e^{ax}$, the result is given by the formula $\frac{x^{n}}{n!} e^{ax}$.
Step 2: Meaning
Here, $a = 2$ and the power of the repeated factor in the denominator is $n = 2$.
Step 3: Analysis
Applying the formula: $\frac{x^{2}}{2!} e^{2x} = \frac{x^{2}}{2} e^{2x}$.
Step 4: Conclusion
This standard formula efficiently solves cases where substituting $D=a$ leads to zero in a repeated factor. Final Answer: (C)
For $P.I. = \frac{1}{(D - a)^{n}} e^{ax}$, the result is given by the formula $\frac{x^{n}}{n!} e^{ax}$.
Step 2: Meaning
Here, $a = 2$ and the power of the repeated factor in the denominator is $n = 2$.
Step 3: Analysis
Applying the formula: $\frac{x^{2}}{2!} e^{2x} = \frac{x^{2}}{2} e^{2x}$.
Step 4: Conclusion
This standard formula efficiently solves cases where substituting $D=a$ leads to zero in a repeated factor. Final Answer: (C)
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