Find the general solution of (2x+1) y'' + 4x y' - 4y = 0 . A…

Find the general solution of $(2x+1)\,y'' + 4x\,y' - 4y = 0$. A polynomial particular solution exists.

Step 1: Find the degree of the polynomial solution

For $y_1=x^n$, the degree-$n$ contribution comes only from $4xy_1'-4y_1$, since $(2x+1)y_1''$ tops out at degree $n-1$: $4x\cdot nx^{n-1} - 4x^n = 4(n-1)x^n.$

This vanishes when $n=1$.

Step 2: Solve for the coefficients and verify

Try $y_1=ax+b$: $y_1'=a$, $y_1''=0$. Substituting, \[(2x+1)(0) + 4x\,a - 4(ax+b) = -4b = 0 \implies b=0.\]

Taking $a=1$ gives $y_1=x$. Check: $(2x+1)(0)+4x(1)-4x=0. $\checkmark$$

Step 3: Substitute $y=y_1v$ and reduce the order

Set $y=xv(x)$: $y'=v+xv'$, $y''=2v'+xv''$. Substituting into $(2x+1)y''+4xy'-4y=0$, $(2x+1)(2v'+xv'') + 4x(v+xv') - 4xv = 0.$

The $4xv$ terms cancel, leaving

\[ x(2x+1)v'' + \bigl[2(2x+1)+4x^2\bigr]v' = 0 \implies x(2x+1)v'' + 2(2x^2+2x+1)v'=0. \]

Step 4: Solve the first-order equation for $w=v'$

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

Dividing $2x^2+2x+1$ by $x(2x+1)=2x^2+x$ leaves remainder $x+1$, so

\[ \frac{2(2x^2+2x+1)}{x(2x+1)} = 2 + \frac{2(x+1)}{x(2x+1)} = 2+\frac2x-\frac2{2x+1}, \] where $\dfrac{2(x+1)}{x(2x+1)}=\dfrac{A}x+\dfrac{B}{2x+1}$ gives $A=2$ (at $x=0$) and $B=-2$ (at $x=-\tfrac12$). Integrating, \[\ln|w| = -2x - 2\ln|x| + \ln|2x+1| + \mathrm{const} \implies w = \frac{A(2x+1)e^{-2x}}{x^2}.\]

Step 5: Integrate for $v$

Write $v' = Ae^{-2x}\!\left(\dfrac2x+\dfrac1{x^2}\right)$. Differentiating by the quotient rule, \[\frac{d}{dx}\!\left(-\frac{e^{-2x}}{x}\right) = \frac{2e^{-2x}}{x}+\frac{e^{-2x}}{x^2} = e^{-2x}\!\left(\frac2x+\frac1{x^2}\right),\] so integration gives

\[ v = -\frac{Ae^{-2x}}{x} + B. \]

Step 6: General solution

Then $y=xv = Bx - Ae^{-2x}$. Renaming $C_1=B$, $C_2=-A$:

Step 7: Verify $y_2=e^{-2x}$

\[ y_2'=-2e^{-2x},\qquad y_2''=4e^{-2x}. \]

\[ (2x+1)(4e^{-2x}) + 4x(-2e^{-2x}) - 4e^{-2x} = e^{-2x}\bigl[4(2x+1)-8x-4\bigr] = 0.\ \checkmark \]

Step 8: Linear independence

\[ W(y_1,y_2) = \begin{vmatrix} x & e^{-2x} \\ 1 & -2e^{-2x}\end{vmatrix} = -2xe^{-2x}-e^{-2x} = -e^{-2x}(2x+1) \neq 0 \quad (x\neq-\tfrac12), \] so $y_1=x$ and $y_2=e^{-2x}$ form a fundamental system.