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: Standard form and verification of $y_1$

Dividing by $x^2(\ln x-1)$ puts the equation in standard form $y''+P(x)y'+Q(x)y=0$ with

\[ P(x) = -\frac{1}{x(\ln x-1)}, \qquad Q(x) = \frac{1}{x^2(\ln x-1)}. \]

With $y_1=x$: $y_1'=1$, $y_1''=0$, and $x^2(\ln x-1)\cdot0 - x\cdot1+x = 0. $\checkmark$

Step 2: Abel's formula shortcut

For $y=y_1v$, the substitution always reduces $y''+Py'+Qy=0$ to $y_1w' + (2y_1'+Py_1)w=0$ with $w=v'$, whose solution is $w = \dfrac{1}{y_1^2}\exp\!\left(-\displaystyle\int P\,dx\right)$ (this is the Wronskian divided by $y_1^2$, by Abel's identity $W'=-PW$).

Step 3: Compute $\int P\,dx$

With $t=\ln x-1$, $dt=dx/x$:

\[ \int P\,dx = -\int\frac{dx}{x(\ln x-1)} = -\int\frac{dt}{t} = -\ln|\ln x-1|. \] So $\exp\!\left(-\int P\,dx\right) = |\ln x-1|$.

Step 4: Compute $w=v'$

\[ w = \frac{1}{y_1^2}\exp\!\left(-\int P\,dx\right) = \frac{\ln x-1}{x^2}, \]

which matches the result obtained by direct substitution in Problem 71 (up to the arbitrary multiplicative constant).

Step 5: Integrate for $v$

Integrating by parts ($u=\ln x-1$, $dq=x^{-2}dx$) as in Problem 71: $v = A\!\int\frac{\ln x-1}{x^2}\,dx = -\frac{A\ln x}{x} + B.$

Step 6: General solution

Then $y = x\cdot v = -A\ln x + Bx$. Renaming $B=C_1$, $-A=C_2$:

Step 7: Verify $y_2=\ln x$

\[ y_2'=\frac1x,\qquad y_2''=-\frac1{x^2}, \]

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

Step 8: Linear independence

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