Question:
\[
\frac{1}{D-a}X=\_
\]
\[
\frac{1}{D-a}X=\_
\]
Show Hint
This general formula works for ANY function $X$, even when specific shortcut methods fail.
Updated On: Jul 4, 2026
- $e^{ax} \int X e^{-ax} dx$
- $e^{-ax} \int X e^{ax} dx$
- $e^{ax} \int \frac{x}{2} e^{ax} dx$
- $\frac{1}{2} e^{ax} \int X e^{-ax} dx$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
This is the general formula for the linear operator $\frac{1}{D - a}$ acting on any function $X$.
Step 2: Meaning
The operation is equivalent to solving the linear differential equation $\frac{dy}{dx} - ay = X$.
Step 3: Analysis
Using the integrating factor $e^{\int -a dx} = e^{-ax}$, the solution is $y e^{-ax} = \int X e^{-ax} dx$.
Step 4: Conclusion
Multiplying both sides by $e^{ax}$ gives $y = e^{ax} \int X e^{-ax} dx$, which is the identity for the operator. Final Answer: (A)
This is the general formula for the linear operator $\frac{1}{D - a}$ acting on any function $X$.
Step 2: Meaning
The operation is equivalent to solving the linear differential equation $\frac{dy}{dx} - ay = X$.
Step 3: Analysis
Using the integrating factor $e^{\int -a dx} = e^{-ax}$, the solution is $y e^{-ax} = \int X e^{-ax} dx$.
Step 4: Conclusion
Multiplying both sides by $e^{ax}$ gives $y = e^{ax} \int X e^{-ax} dx$, which is the identity for the operator. Final Answer: (A)
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