Question:

The general solution of (D2+6D+9)y=e3xx2(D^2+6D+9)y=\frac{e^{-3x}}{x^2}, where DddxD\equiv \frac{d}{dx} is
(given that c1 and c2 are arbitrary constants)

Updated On: Mar 12, 2025
  • y=(c1+c2x)e3x+e3x2xy=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}
  • y=(c1+c2x)e3x+e3x2xy=(c_1+c_2x)e^{3x}+\frac{e^{3x}}{2x}
  • y=(c1+c2x2)e3x+e3x2xy=(c_1+c_2x^2)e^{-3x}+\frac{e^{-3x}}{2x}
  • y=(c1+c2x)e3x+e3x2xy=(c_1+c_2x)e^{-3x}+\frac{e^{3x}}{2x}
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The Correct Option is A

Solution and Explanation

The correct answer is(A): y=(c1+c2x)e3x+e3x2xy=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}
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