Solve by variation of parameters y'' - 6y' + 9y =…

Solve by variation of parameters $y'' - 6y' + 9y = \frac{9x^2+6x+2}{x^3}$.

Step 1: Homogeneous equation

The characteristic equation $r^2-6r+9=(r-3)^2=0$ has the double root $r=3$, so

\[ y_1=e^{3x}, \qquad y_2=xe^{3x}. \]

Step 2: Wronskian

\[ W(y_1,y_2)=\begin{vmatrix}e^{3x} & xe^{3x}\\3e^{3x} & (1+3x)e^{3x}\end{vmatrix} = e^{6x}\bigl[(1+3x)-3x\bigr] = e^{6x}. \]

Step 3: Variation-of-parameters system

With $f(x)=\dfrac{9x^2+6x+2}{x^3}$ and $W=e^{6x}$: \[u_1' = -\frac{y_2f}{W} = -\frac{e^{-3x}(9x^2+6x+2)}{x^2}, \qquad u_2' = \frac{y_1f}{W} = \frac{e^{-3x}(9x^2+6x+2)}{x^3}.\]

Step 4: Integrate for $u_1$

\[ u_1 = -\int e^{-3x}\!\left(9+\frac6x+\frac2{x^2}\right)dx = -9\!\int e^{-3x}dx - 6\!\int\frac{e^{-3x}}{x}dx - 2\!\int\frac{e^{-3x}}{x^2}dx. \]

Integrating $\int e^{-3x}x^{-2}dx$ by parts ($U=e^{-3x}$, $dV=x^{-2}dx$): \[\int\frac{e^{-3x}}{x^2}\,dx = -\frac{e^{-3x}}{x} - 3\!\int\frac{e^{-3x}}{x}\,dx.\]

Substituting back, the $\int e^{-3x}/x\,dx$ (exponential-integral) terms cancel exactly, $-6+6=0$: \[u_1 = -9\!\int e^{-3x}dx - 6\!\int\frac{e^{-3x}}{x}dx + \frac{2e^{-3x}}{x}+6\!\int\frac{e^{-3x}}{x}dx = 3e^{-3x}+\frac{2e^{-3x}}{x} = e^{-3x}\!\left(3+\frac2x\right).\]

Step 5: Integrate for $u_2$

\[ u_2 = \int e^{-3x}\!\left(\frac9x+\frac6{x^2}+\frac2{x^3}\right)dx. \]

Using the Step 4 result for $\int e^{-3x}/x^2\,dx$, and integrating $\int e^{-3x}x^{-3}dx$ by parts ($U=e^{-3x}$, $dV=x^{-3}dx$): \[\int\frac{e^{-3x}}{x^3}\,dx = -\frac{e^{-3x}}{2x^2} - \frac32\!\int\frac{e^{-3x}}{x^2}dx = -\frac{e^{-3x}}{2x^2}+\frac{3e^{-3x}}{2x}+\frac92\!\int\frac{e^{-3x}}{x}dx.\]

Collecting the coefficients of $\int e^{-3x}/x\,dx$ across all three pieces of $u_2$: $9-18+9=0$, so that term cancels exactly, leaving

\[ u_2 = -\frac{6e^{-3x}}{x}-\frac{e^{-3x}}{x^2}+\frac{3e^{-3x}}{x} = -\frac{3e^{-3x}}{x}-\frac{e^{-3x}}{x^2} = -e^{-3x}\!\left(\frac3x+\frac1{x^2}\right). \]

(Both cancellations were checked independently by differentiating the closed forms of $u_1,u_2$ and confirming they reproduce $u_1',u_2'$ from Step 3 exactly.)

Step 6: Particular solution

\[ y^* = u_1e^{3x}+u_2xe^{3x} = \left(3+\frac2x\right) - x\!\left(\frac3x+\frac1{x^2}\right) = 3+\frac2x-3-\frac1x = \frac1x. \]

Step 7: Verify $y^*$

With $y^*=1/x$: $y^{*\prime}=-1/x^2$, $y^{*\prime\prime}=2/x^3$, so

\[ y^{*\prime\prime}-6y^{*\prime}+9y^* = \frac2{x^3}+\frac6{x^2}+\frac9x = \frac{2+6x+9x^2}{x^3} = \frac{9x^2+6x+2}{x^3}.\ \checkmark \]

Step 8: General solution

The solution is recorded in the result below.