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 ____.

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

Hint:

Use the exponential shift \(y=e^{-3x}u(x)\) to reduce the equation to \(u''=1/x^3\), then integrate twice.

Answer 1

The Correct Option is C

Step 1: Write the equation as $(D+3)^2y=\frac{e^{-3x}}{x^3}$, since $D^2+6D+9=(D+3)^2$.

Step 2: Use the exponential shift: set $y=e^{-3x}u(x)$. Then $Dy=e^{-3x}(u'-3u)$ and $D^2y=e^{-3x}(u''-6u'+9u)$.

Step 3: Substitute into $(D^2+6D+9)y$: \[e^{-3x}(u''-6u'+9u)+6e^{-3x}(u'-3u)+9e^{-3x}u=e^{-3x}u''.\] So the equation reduces to $e^{-3x}u''=\frac{e^{-3x}}{x^3}$, that is $u''=\frac{1}{x^3}$.

Step 4: Integrate twice: \[u'=-\frac{1}{2x^2}+c_2,\qquad u=\frac{1}{2x}+c_2x+c_1.\]

Step 5: Then $y=e^{-3x}u=(c_1+c_2x)e^{-3x}+\frac{e^{-3x}}{2x}$, the same general solution obtained by an algebraic reduction instead of the Wronskian formulas.

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