Question:
The general solution of the ODE \((D^2+6D+9)y=\frac{e^{-3x}}{x^3}\), \(D=\frac{d}{dx}\) (\(c_1,c_2\) are arbitrary constants) is ____.
The general solution of the ODE \((D^2+6D+9)y=\frac{e^{-3x}}{x^3}\), \(D=\frac{d}{dx}\) (\(c_1,c_2\) are arbitrary constants) is ____.
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Use the exponential shift \(y=e^{-3x}u(x)\) to reduce the equation to \(u''=1/x^3\), then integrate twice.
Updated On: Jul 3, 2026
- \(y(x)=(c_1+c_2x)e^{3x}+\frac{e^{3x}}{2x}\)
- \(y(x)=(c_1+c_2x)e^{-3x}+\frac{e^{3x}}{2x}\)
- \(y(x)=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}\)
- \(y(x)=(c_1+c_2x)e^{3x}+\frac{e^{-3x}}{2x}\)
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The Correct Option is C
Solution and Explanation
Step 1: The auxiliary equation for \((D^2+6D+9)y=0\) is \(m^2+6m+9=0\), which factors as \((m+3)^2=0\), giving a repeated root \(m=-3\).
Step 2: The complementary function is \(y_c=(c_1+c_2x)e^{-3x}\).
Step 3: Since the forcing term \(\frac{e^{-3x}}{x^3}\) is not a polynomial times exponential, use variation of parameters with \(y_1=e^{-3x}\) and \(y_2=xe^{-3x}\). The Wronskian is \[W=y_1y_2'-y_1'y_2=e^{-6x}.\]
Step 4: With \(f(x)=\frac{e^{-3x}}{x^3}\), compute \[u_1'=-\frac{y_2f}{W}=-\frac{1}{x^2},\qquad u_2'=\frac{y_1f}{W}=\frac{1}{x^3}.\]
Step 5: Integrate: \(u_1=\frac{1}{x}\) and \(u_2=-\frac{1}{2x^2}\).
Step 6: The particular solution is \[y_p=u_1y_1+u_2y_2=\frac{e^{-3x}}{x}-\frac{e^{-3x}}{2x}=\frac{e^{-3x}}{2x}.\]
\[\boxed{y(x)=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}}\]
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