Find the general solution of y'' + (x)/(1-x) y' - (1)/(1-x)…

Find the general solution of $y'' + \frac{x}{1-x}\,y' - \frac{1}{1-x}\,y = 0$, 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: \[0 + \frac{x}{1-x}\cdot 1 - \frac{1}{1-x}\cdot x = \frac{x}{1-x}-\frac{x}{1-x} = 0.\ \checkmark\]

Step 2: Substitution $y=xv$

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

Step 3: Reduce the order

Substituting: $(2v'+xv'') + \frac{x}{1-x}(v+xv') - \frac{x}{1-x}v = 0.$

The $\dfrac{xv}{1-x}$ terms cancel, leaving: \[xv'' + v'\!\left(2+\frac{x^2}{1-x}\right) = 0 \implies xv'' + v'\cdot\frac{x^2-2x+2}{1-x} = 0.\]

Step 4: Solve for $w=v'$

\[ xw' + \frac{x^2-2x+2}{1-x}\,w = 0 \implies \frac{dw}{w} = -\frac{x^2-2x+2}{x(1-x)}\,dx. \]

Polynomial division followed by partial fractions gives: \[\frac{x^2-2x+2}{x(1-x)} = -1 + \frac{2}{x} + \frac{1}{1-x}.\] So: \[\ln|w| = x - 2\ln|x| + \ln|1-x| + \mathrm{const} \implies w = \frac{A(1-x)e^x}{x^2}.\]

Step 5: Integrate for $v$

Write $v' = Ae^x\!\left(\dfrac1{x^2}-\dfrac1x\right)$. Since $\displaystyle\frac{d}{dx}\!\left(-\frac{e^x}{x}\right) = -\frac{e^x}{x}+\frac{e^x}{x^2} = e^x\!\left(\frac1{x^2}-\frac1x\right)$, integration is immediate:

\[ v = -\frac{Ae^x}{x} + B. \]

Step 6: General solution

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

Step 7: Verify $y_2=e^x$

\[ e^x + \frac{x}{1-x}e^x - \frac{1}{1-x}e^x = e^x\cdot\frac{(1-x)+x-1}{1-x} = 0.\ \checkmark \]

Step 8: Linear independence

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