Find the general solution of x(2x+1) y'' + 2(x+1) y' - 2y = 0

Find the general solution of $x(2x+1)\,y'' + 2(x+1)\,y' - 2y = 0$.

Step 1: Finding a Polynomial Solution

Try $y_1=ax+b$, so $y_1'=a$, $y_1''=0$. Substituting: \[0 + 2(x+1)a - 2(ax+b) = 2a-2b = 0 \implies b=a.\]

Choosing $a=1$: $y_1=x+1$.

Step 2: Verify $y_1$

\[ 0 + 2(x+1)\cdot1 - 2(x+1) = 0.\ \checkmark \]

Step 3: Substitution $y=(x+1)v$

\[ y' = v+(x+1)v', \qquad y'' = 2v'+(x+1)v''. \]

Step 4: Reduce the order

Substituting into $x(2x+1)y''+2(x+1)y'-2y=0$, the $v$ terms cancel, leaving: $x(2x+1)(x+1)v'' + \bigl[2x(2x+1)+2(x+1)^2\bigr]v' = 0.$

Since $2x(2x+1)+2(x+1)^2 = 4x^2+2x+2x^2+4x+2 = 6x^2+6x+2$: $x(2x+1)(x+1)\,v'' + (6x^2+6x+2)\,v' = 0.$

Step 5: Solve for $w=v'$

Separating variables and decomposing $\dfrac{6x^2+6x+2}{x(2x+1)(x+1)}=\dfrac2x-\dfrac{2}{2x+1}+\dfrac{2}{x+1}$: \[\ln|w| = -2\ln|x|+\ln|2x+1|-2\ln|x+1|+\mathrm{const} \implies w = A\,\frac{2x+1}{x^2(x+1)^2}.\]

Step 6: Integrate for $v$

Since $\dfrac{2x+1}{x^2(x+1)^2}=\dfrac{1}{x^2}-\dfrac{1}{(x+1)^2}$: $v = A\left(-\frac1x+\frac1{x+1}\right)+B.$

Step 7: General solution

Multiply through by $y_1=x+1$, combining the $A$-bracket into a single fraction rather than distributing term-by-term: \[(x+1)\!\left(-\frac1x+\frac1{x+1}\right) = -\frac{x+1}{x}+1 = \frac{-(x+1)+x}{x} = -\frac1x.\] So $y=(x+1)v=-\dfrac{A}{x}+B(x+1)$. Renaming $C_1=B$, $C_2=-A$:

Step 8: Verify $y_2=1/x$

\[ y_2'=-\frac1{x^2},\qquad y_2''=\frac2{x^3}. \]

\[ x(2x+1)\cdot\frac2{x^3}+2(x+1)\!\left(-\frac1{x^2}\right)-\frac2x = \frac{2(2x+1)-2(x+1)}{x^2}-\frac2x = \frac{2x}{x^2}-\frac2x = 0.\ \checkmark \]

Step 9: Linear independence

\[ W(y_1,y_2) = (x+1)\!\left(-\frac1{x^2}\right) - \frac1x\cdot1 = -\frac{x+1}{x^2}-\frac1x = -\frac{2x+1}{x^2} \neq 0, \]

confirming $y_1=x+1$ and $y_2=1/x$ form a fundamental system for $x\neq0$.