Find the general solution of x^2ln x· y'' - x y' + y = 0 . A…

Find the general solution of $x^2\ln x\cdot y'' - x\,y' + y = 0$. A polynomial particular solution exists.

Step 1: Find the particular solution by matching

Try $y_1=x^n$: $y_1'=nx^{n-1}$, $y_1''=n(n-1)x^{n-2}$. Substituting, $x^2\ln x\cdot n(n-1)x^{n-2} - x\cdot nx^{n-1} + x^n = x^n\bigl[n(n-1)\ln x + (1-n)\bigr] = 0.$

The coefficient of $\ln x$ is $n(n-1)$, and $\ln x$ is not constant, so this can vanish identically only if both $n(n-1)=0$ and $1-n=0$ hold together — which forces $n=1$ (the other root $n=0$ fails the second condition). Check directly with $y_1=x$:

\[ x^2\ln x\cdot(0) - x\cdot(1) + x = 0.\ \checkmark \]

Step 2: Substitute $y=y_1v$ and reduce the order

Set $y=xv(x)$: $y'=v+xv'$, $y''=2v'+xv''$. Substituting into $x^2\ln x\cdot y''-xy'+y=0$, the $xv$ terms cancel and, dividing by $x^2$, $x\ln x\cdot v'' + (2\ln x-1)v' = 0.$

Step 3: Solve the first-order equation for $w=v'$

\[ \frac{dw}{w} = -\frac{2\ln x-1}{x\ln x}\,dx = \left(-\frac2x+\frac1{x\ln x}\right)dx. \]

Integrating the second term with $t=\ln x$, $dt=dx/x$: \[\ln|w| = -2\ln x + \ln|\ln x| + \mathrm{const} \implies w = \frac{A\ln x}{x^2}.\]

Step 4: Integrate for $v$

Integrate by parts with $u=\ln x$, $dq=x^{-2}dx$ (so $du=dx/x$, $q=-1/x$): \[\int\frac{\ln x}{x^2}\,dx = -\frac{\ln x}{x} + \int\frac{dx}{x^2} = -\frac{\ln x}{x}-\frac1x + \mathrm{const}.\] So $v = -\frac{A(\ln x+1)}{x} + B.$

Step 5: General solution

Then $y=xv = -A(\ln x+1) + Bx$. The two independent pieces are the original $y_1=x$ (coefficient $B$) and a genuinely new solution $\ln x+1$ (coefficient $-A$); renaming $C_1=B$, $C_2=-A$:

Step 6: Verify $y_2=\ln x+1$

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

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

Step 7: Linear independence

\[ W(y_1,y_2) = \begin{vmatrix} x & \ln x+1 \\ 1 & 1/x \end{vmatrix} = 1-(\ln x+1) = -\ln x \neq 0 \quad (x\neq1), \] so $y_1=x$ and $y_2=\ln x+1$ form a fundamental system.