Determine whether y = (1+ ln x)/(x - x ln x) satisfies

Determine whether $y = \dfrac{1+\ln x}{x - x\ln x}$ satisfies the differential equation $2x^{2}y' - x^{2}y^{2} - 1 = 0$.

Step 1: Factor the denominator

\[ y = \frac{1+\ln x}{x\bigl(1-\ln x\bigr)}. \]

Step 2: Differentiate by the quotient rule

With $u = 1+\ln x$, $u' = \dfrac1x$, and $v = x - x\ln x$, whose derivative is

\[ v' = 1 - \bigl(\ln x + 1\bigr) = -\ln x, \]

the numerator of $y'$ becomes

\[ u'v - uv' = \frac1x\cdot x\bigl(1-\ln x\bigr) - \bigl(1+\ln x\bigr)\bigl(-\ln x\bigr) = \bigl(1-\ln x\bigr) + \ln x + \ln^{2}x = 1+\ln^{2}x. \]

The two $\ln x$ terms cancel, leaving

\[ y' = \frac{1+\ln^{2}x}{x^{2}\bigl(1-\ln x\bigr)^{2}}. \]

Step 3: Compute $y^{2}$

\[ y^{2} = \frac{\bigl(1+\ln x\bigr)^{2}}{x^{2}\bigl(1-\ln x\bigr)^{2}}. \]

Step 4: Substitute into the equation

The factors $x^{2}$ cancel against the $x^{2}$ in front of each term: \[2x^{2}y' - x^{2}y^{2} - 1 = \frac{2\bigl(1+\ln^{2}x\bigr) - \bigl(1+\ln x\bigr)^{2}}{\bigl(1-\ln x\bigr)^{2}} - 1.\]

Step 5: Expand the numerator

It turns out to be exactly the denominator: $2 + 2\ln^{2}x - 1 - 2\ln x - \ln^{2}x = 1 - 2\ln x + \ln^{2}x = \bigl(1-\ln x\bigr)^{2}$.

Therefore \[2x^{2}y' - x^{2}y^{2} - 1 = \frac{\bigl(1-\ln x\bigr)^{2}}{\bigl(1-\ln x\bigr)^{2}} - 1 = 1-1 = 0.\]

Step 6: Verification

\[x = 1:\quad \ln 1 = 0 \;\Longrightarrow\; y = 1,\quad y' = \frac{1+0}{1\cdot 1} = 1\]

\[2\cdot 1\cdot 1 - 1\cdot 1 - 1 = 2 - 1 - 1 = 0 \;\;\checkmark\]