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): \(y=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}\)