Solve by variation of parameters x^3(y''-y) = x^2-2

Solve by variation of parameters $x^3(y''-y) = x^2-2$.

Step 1: Standard form

\[ y''-y = \frac{x^2-2}{x^3} = \frac1x-\frac2{x^3}. \]

Step 2: Homogeneous equation

The characteristic equation $r^2-1=0$ gives $r=\pm1$, so

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

Step 3: Wronskian

\[ W(y_1,y_2)=\begin{vmatrix}e^x & e^{-x}\\e^x & -e^{-x}\end{vmatrix} = -1-1=-2. \]

Step 4: Variation-of-parameters system

With $f(x)=\dfrac1x-\dfrac2{x^3}$ and $W=-2$, Cramer's rule gives

\[ u_1' = -\frac{y_2f}{W} = \frac{e^{-x}}{2}\!\left(\frac1x-\frac2{x^3}\right), \qquad u_2' = \frac{y_1f}{W} = -\frac{e^x}{2}\!\left(\frac1x-\frac2{x^3}\right). \]

Step 5: Integrate for $u_1$

\[ u_1 = \frac12\int\frac{e^{-x}}{x}\,dx - \int\frac{e^{-x}}{x^3}\,dx. \]

Reduce $\int e^{-x}x^{-3}dx$ by two integrations by parts, each time peeling off one power of $x$: \[\int\frac{e^{-x}}{x^3}\,dx = -\frac{e^{-x}}{2x^2} - \frac12\int\frac{e^{-x}}{x^2}\,dx, \qquad \int\frac{e^{-x}}{x^2}\,dx = -\frac{e^{-x}}{x} - \int\frac{e^{-x}}{x}\,dx.\]

Substituting the second into the first, \[\int\frac{e^{-x}}{x^3}\,dx = -\frac{e^{-x}}{2x^2}+\frac{e^{-x}}{2x}+\frac12\int\frac{e^{-x}}{x}\,dx,\]

and the non-elementary term $\int e^{-x}/x\,dx$ (an exponential-integral $\mathrm{Ei}$ function) cancels exactly against the identical term already present in $u_1$: \[u_1 = \frac12\!\int\!\frac{e^{-x}}{x}dx - \left(-\frac{e^{-x}}{2x^2}+\frac{e^{-x}}{2x}+\frac12\!\int\!\frac{e^{-x}}{x}dx\right) = \frac{e^{-x}}{2x^2}-\frac{e^{-x}}{2x} = \frac{e^{-x}(1-x)}{2x^2}.\]

Step 6: Integrate for $u_2$

\[ u_2 = -\frac12\int\frac{e^x}{x}\,dx + \int\frac{e^x}{x^3}\,dx. \]

By the same two-step reduction (now with $e^x$): \[\int\frac{e^x}{x^3}\,dx = -\frac{e^x}{2x^2}+\frac12\int\frac{e^x}{x^2}\,dx, \qquad \int\frac{e^x}{x^2}\,dx = -\frac{e^x}{x}+\int\frac{e^x}{x}\,dx,\]

\[ \int\frac{e^x}{x^3}\,dx = -\frac{e^x}{2x^2}-\frac{e^x}{2x}+\frac12\int\frac{e^x}{x}\,dx. \]

Again the $\int e^x/x\,dx$ terms cancel exactly: \[u_2 = -\frac12\!\int\!\frac{e^x}{x}dx + \left(-\frac{e^x}{2x^2}-\frac{e^x}{2x}+\frac12\!\int\!\frac{e^x}{x}dx\right) = -\frac{e^x}{2x^2}-\frac{e^x}{2x} = -\frac{e^x(x+1)}{2x^2}.\]

(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 4 exactly.)

Step 7: Particular solution

\[ y^* = u_1e^x+u_2e^{-x} = \frac{1-x}{2x^2}-\frac{x+1}{2x^2} = \frac{-2x}{2x^2} = -\frac1x. \]

Step 8: Verify $y^*$

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

\[ y^{*\prime\prime}-y^* = -\frac2{x^3}+\frac1x = \frac{-2+x^2}{x^3} = \frac{x^2-2}{x^3}.\ \checkmark \]

Step 9: General solution

The solution is recorded in the result below.