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