Question

A solution of the differential equation $(D^2 - 1)y = 2^x + e^{-x}; D = \frac{d}{dx}$ is

• A. $y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{1 - (\log 2)^2} 2^x - \frac{1}{2} x^2 e^{-x}$
• B. $y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{1 - (\log 2)^2} 2^x - \frac{1}{2} e^{-x}$
• C. $y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{1 - (\log 2)^2} 2^x - \frac{x}{2} e^{-x}$
• D. $y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{(\log 2)^2 - 1} 2^x - \frac{x}{2} e^{-x}$

Hint:

Case of failure rule: When evaluating $\frac{1}{f(D)} e^{a x}$ and $f(a) = 0$, differentiate denominator with respect to $D$ and multiply numerator by $x$: $\frac{1}{f(D)} e^{a x} = x \frac{1}{f'(D)} e^{a x}$. Here $x \frac{1}{2D} e^{-x} = \frac{x e^{-x}}{2(-1)} = -\frac{x}{2} e^{-x}$.

Answer 1

The Correct Option is D

Step 1: Concept
The general solution to a linear non-homogeneous differential equation $(D^2 - 1)y = f(x)$ is $y = y_c + y_p$, where $y_c$ is the complementary function and $y_p$ is the particular integral.

Step 2: Key Formulas and Approach
1. Complementary Function ($y_c$): Solve auxiliary equation $m^2 - 1 = 0$.
2. Particular Integral ($y_p$):
- For $2^x = e^{x \ln 2}$, use $\frac{1}{f(D)} e^{a x} = \frac{1}{f(a)} e^{a x}$ where $a = \log 2$.
- For $e^{-x}$, since $f(-1) = (-1)^2 - 1 = 0$ (case of failure), use $\frac{1}{D - a} e^{a x} = x e^{a x}$.

Step 3: Step-by-step Explanation

• 1. Finding Complementary Function $y_c$: Auxiliary equation: $m^2 - 1 = 0 \implies m = 1, -1$. \[ y_c = C_1 e^x + C_2 e^{-x} \]
• 2. Finding Particular Integral $y_{p_1$ for $2^x$:} Rewrite $2^x = e^{x \ln 2}$. Replace $D$ by $\log 2$: \[ y_{p_1} = \frac{1}{D^2 - 1} 2^x = \frac{1}{(\log 2)^2 - 1} 2^x \]
• 3. Finding Particular Integral $y_{p_2$ for $e^{-x}$:} Factor $D^2 - 1 = (D - 1)(D + 1)$: \[ y_{p_2} = \frac{1}{(D - 1)(D + 1)} e^{-x} = \frac{1}{-1 - 1} \cdot \left( \frac{1}{D + 1} e^{-x} \right) \] \[ = -\frac{1}{2} \left( x e^{-x} \right) = -\frac{x}{2} e^{-x} \]
• 4. Combining terms for complete general solution: \[ y(x) = y_c + y_{p_1} + y_{p_2} \] \[ y(x) = C_1 e^x + C_2 e^{-x} + \frac{1}{(\log 2)^2 - 1} 2^x - \frac{x}{2} e^{-x} \]

Step 4: Final Answer
The general solution matches Option (D). Thus, Option (D) is correct.