Question

The general solution of $(D^2+6D+9)y=\frac{e^{-3x}}{x^2}$, where $D\equiv \frac{d}{dx}$ is
(given that c₁ and c₂ are arbitrary constants)

• A. $y=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}$
• B. $y=(c_1+c_2x)e^{3x}+\frac{e^{3x}}{2x}$
• C. $y=(c_1+c_2x^2)e^{-3x}+\frac{e^{-3x}}{2x}$
• D. $y=(c_1+c_2x)e^{-3x}+\frac{e^{3x}}{2x}$

Answer 1

The Correct Option is A

The correct answer is(A): $y=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}$