Find a particular solution of the homogeneous part of (*)…

Find a particular solution of the homogeneous part of $(*)\!: \qquad x^4y'' + (x^2-1)xy' - (x^2-1)y = \frac{1}{x^5}$, then solve the full equation.

Step 1: A power solution of the homogeneous equation

The homogeneous equation is $(H)\!:\ x^4y''+x(x^2-1)y'-(x^2-1)y=0$. Try $y=x^k$, so $y'=kx^{k-1}$, $y''=k(k-1)x^{k-2}$. Substituting: $k(k-1)x^{k+2} + k(x^2-1)x^k - (x^2-1)x^k = 0$.

Collecting terms, this factors cleanly: $(k-1)x^k\bigl[(k+1)x^2-1\bigr] = 0$.

Since the bracket $(k+1)x^2-1$ cannot vanish identically in $x$ for any fixed $k$, the equation holds for all $x$ only if the prefactor vanishes: $k=1$. Hence

\[ y_1 = x. \]

Check: $y_1'=1$, $y_1''=0$, so $x^4\cdot0 + x(x^2-1)\cdot1 - (x^2-1)\cdot x = x(x^2-1)-x(x^2-1)=0.\ \checkmark$

Step 2: Reduction of order: $y=xv$

With $y=xv$: $y'=v+xv'$, $y''=2v'+xv''$. Substituting into $(H)$, the terms in $v$ alone (those not involving $v'$ or $v''$) cancel exactly — a consequence of $y_1=x$ already solving $(H)$ — leaving a second-order equation with no $v$ term:

\[ x^5v'' + x^2(3x^2-1)v' = 0 \;\Longrightarrow\; x^3v'' + (3x^2-1)v' = 0. \]

Step 3: Solving for $w=v'$

Setting $w=v'$ turns this into a first-order separable equation, $x^3w'+(3x^2-1)w=0$: \[\frac{dw}{w} = \frac{1-3x^2}{x^3}\,dx = \left(\frac{1}{x^3}-\frac3x\right)dx.\]

Integrating, \[\ln|w| = -\frac{1}{2x^2} - 3\ln|x| + \mathrm{const} \;\Longrightarrow\; w = \frac{A}{x^3}\,e^{-1/(2x^2)}.\]

Step 4: Integrating $v'=w$

With the substitution $t=1/(2x^2)$, $dt=-dx/x^3$: \[v = \int\frac{A}{x^3}e^{-1/(2x^2)}\,dx = A\int e^{-t}(-dt) = -Ae^{-t}+B = -Ae^{-1/(2x^2)}+B.\]

Choosing $A=-1,\,B=0$ (any convenient nontrivial member of this family) gives $v=e^{-1/(2x^2)}$, and therefore a second homogeneous solution

\[ y_2 = x\,e^{-1/(2x^2)}. \]

Step 5: Verification of $y_2$ and the Wronskian

Differentiating: $y_2' = e^{-1/(2x^2)}\bigl(1+\tfrac1{x^2}\bigr)$, and

\[ y_2'' = e^{-1/(2x^2)}\left(\frac{1}{x^5}-\frac{1}{x^3}\right). \]

Substituting $y_2,y_2',y_2''$ into $(H)=x^4y''+x(x^2-1)y'-(x^2-1)y$ and simplifying term by term shows every power of $x$ cancels, leaving $0.\ \checkmark$ The Wronskian is

\[ W = y_1y_2' - y_1'y_2 = x\,e^{-1/(2x^2)}\Bigl(1+\frac1{x^2}\Bigr) - e^{-1/(2x^2)}\cdot x = \frac{e^{-1/(2x^2)}}{x}, \]

nonzero for $x\neq0$, confirming $y_1,y_2$ are linearly independent.

Step 6: Standard form and variation of parameters

Dividing $(*)$ by $x^4$ puts it in standard form $y''+p(x)y'+q(x)y=f(x)$ with forcing term

\[ f(x) = \frac{1}{x^9}. \]

Variation of parameters seeks $y_p=-y_1A+y_2B$ with $A=\displaystyle\int\frac{y_2f}{W}\,dx$, $B=\displaystyle\int\frac{y_1f}{W}\,dx$. Since $y_2f/W = x^{-8}e^{-1/(2x^2)}\cdot xe^{1/(2x^2)} = x^{-7}$:

\[ A = \int x^{-7}\,dx = -\frac{1}{6x^6}. \]

And $y_1f/W = x^{-8}\cdot xe^{1/(2x^2)} = x^{-7}e^{1/(2x^2)}$; with $t=1/(2x^2)$ as before (so $x^{-7}dx=-4t^2\,dt$) and $\int t^2e^t\,dt = e^t(t^2-2t+2)$: \[B = \int x^{-7}e^{1/(2x^2)}\,dx = -4\int t^2e^t\,dt = -4e^t(t^2-2t+2) = -e^{1/(2x^2)}\left(\frac{1}{x^4}-\frac{4}{x^2}+8\right).\]

Step 7: Assembling the particular solution

\[ y_p = -y_1A + y_2B = -x\left(-\frac{1}{6x^6}\right) + x\,e^{-1/(2x^2)}\left[-e^{1/(2x^2)}\left(\frac{1}{x^4}-\frac{4}{x^2}+8\right)\right]. \]

The exponentials cancel in the second term, leaving

\[ y_p = \frac{1}{6x^5} - x\left(\frac{1}{x^4}-\frac{4}{x^2}+8\right) = \frac{1}{6x^5} - \frac{1}{x^3} + \frac{4}{x} - 8x. \]

Since $-8x$ is itself a multiple of the homogeneous solution $y_1=x$, it can be absorbed into the arbitrary constant $C_1$ of the general solution (redefine $C_1\to C_1-8$), giving the cleaner equivalent particular solution $1/(6x^5)-1/x^3+4/x$.

Step 8: Verification of the full solution

Writing the general solution as $y=C_1x+C_2xe^{-1/(2x^2)}+\tfrac{1}{6x^5}-\tfrac{1}{x^3}+\tfrac{4}{x}$ and substituting into $x^4y''+(x^2-1)xy'-(x^2-1)y$: the $C_1x$ and $C_2xe^{-1/(2x^2)}$ terms contribute $0$ by Steps 1 and 5, and the remaining fixed terms $\tfrac{1}{6x^5}-\tfrac{1}{x^3}+\tfrac{4}{x}$ reproduce exactly $1/x^5$, matching the right-hand side of $(*)$ identically in $C_1,C_2.\ \checkmark$

\[y = C_1x + C_2x\,e^{-1/(2x^2)} + \frac{1}{6x^5} - \frac{1}{x^3} + \frac{4}{x}\]