Find the general solution of x^2(ln x - 1) y'' - x y' + y =…

Find the general solution of $x^2(\ln x - 1)\,y'' - x\,y' + y = 0$, given that $y_1 = x$ is a particular solution.

Step 1: Verify $y_1$

With $y_1=x$: $y_1'=1$, $y_1''=0$. Substituting: $x^2(\ln x-1)\cdot 0 - x\cdot 1 + x = -x+x = 0.$ $\checkmark$

Step 2: Substitution $y=xv$

\[ y' = v + xv', \qquad y'' = 2v' + xv''. \]

Step 3: Reduce the order

Substituting into the equation: $x^2(\ln x-1)(2v'+xv'') - x(v+xv') + xv = 0.$

The $-xv$ and $+xv$ terms cancel, leaving: \[x^3(\ln x-1)\,v'' + x^2\bigl[2(\ln x-1)-1\bigr]v' = 0 \implies x^3(\ln x-1)v'' + x^2(2\ln x-3)v' = 0.\]

Step 4: Solve for $w=v'$

Divide by $x^3(\ln x-1)$ and set $w=v'$: $w' + \frac{2\ln x-3}{x(\ln x-1)}\,w = 0.$

Write $2\ln x-3 = 2(\ln x-1)-1$, so $\dfrac{2\ln x-3}{x(\ln x-1)} = \dfrac{2}{x}-\dfrac{1}{x(\ln x-1)}$. Separating and integrating (the second term via $t=\ln x-1$): \[\ln|w| = -2\ln x + \ln|\ln x-1| + \mathrm{const} \implies w = \frac{A(\ln x-1)}{x^2}.\]

Step 5: Integrate for $v$

Integrate by parts with $u=\ln x-1$, $dq=x^{-2}dx$, so $du=dx/x$, $q=-1/x$: \[v = A\!\int\frac{\ln x-1}{x^2}\,dx = A\left[-\frac{\ln x-1}{x} - \frac{1}{x}\right] + B = -\frac{A\ln x}{x} + B.\]

Step 6: General solution

Then $y = x\!\left(-\dfrac{A\ln x}{x}+B\right) = -A\ln x + Bx$. Renaming $B=C_1$, $-A=C_2$:

Step 7: Verify $y_2=\ln x$

\[ y_2'=\frac1x,\qquad y_2''=-\frac{1}{x^2}. \]

\[ x^2(\ln x-1)\!\left(-\frac1{x^2}\right) - x\cdot\frac1x + \ln x = -(\ln x-1)-1+\ln x = 0.\ \checkmark \]

Step 8: Linear independence

\[ W(y_1,y_2) = \begin{vmatrix} x & \ln x \\ 1 & 1/x \end{vmatrix} = 1-\ln x \neq 0 \quad (x\neq e), \] so $y_1,y_2$ form a fundamental system.