The general solution of ((D^2+6D+9)y=\frac{e^{-3x}}{x^2}), w... | Mathematics | SolveeIt | Solveeit

Today, August 24

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

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

Explanation

Solution

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