Find the general solution of x^2(2x-1) y''' + (4x-3)x y''…

Find the general solution of $x^2(2x-1)\,y''' + (4x-3)x\,y'' - 2x\,y' + 2y = 0$, given that $y_1 = x$ and $y_2 = \dfrac1x$ are particular solutions.

Step 1: Verify $y_1$ and $y_2$

For $y_1=x$: $y_1'=1$, $y_1''=y_1'''=0$, so $-2x(1)+2x=0. $\checkmark$$

For $y_2=x^{-1}$: $y_2'=-x^{-2}$, $y_2''=2x^{-3}$, $y_2'''=-6x^{-4}$. Substituting, \[\frac{-6(2x-1)}{x^2}+\frac{2(4x-3)}{x^2}+\frac{2}{x}+\frac2x = \frac{-12x+6+8x-6}{x^2}+\frac4x = -\frac4x+\frac4x = 0.\ \checkmark\]

Step 2: First reduction: $y=xu$

Since $y_1=x$, set $y=xu(x)$, so $y'=u+xu'$, $y''=2u'+xu''$, $y'''=3u''+xu'''$. Substituting, $x^2(2x-1)(3u''+xu''') + x(4x-3)(2u'+xu'') - 2x(u+xu') + 2xu = 0.$

The $u$ terms cancel ($-2xu+2xu=0$); the remaining terms collect to

\[ x^3(2x-1)u''' + 2x^2(5x-3)u'' + 6x(x-1)u' = 0, \]

and, dividing by $x$: $x^2(2x-1)u''' + 2x(5x-3)u'' + 6(x-1)u' = 0.$

Step 3: Second reduction: $w=u'$

\[ x^2(2x-1)w'' + 2x(5x-3)w' + 6(x-1)w = 0. \]

Since $y_2=1/x=x\cdot u_2$ gives $u_2=x^{-2}$, the known solution $y_2$ corresponds to $u_2'=-2x^{-3}$ — so $w_1=x^{-3}$ (up to a constant) should solve this equation. Checking directly, with $w=x^{-3}$, $w'=-3x^{-4}$, $w''=12x^{-5}$:

\[ \frac{12(2x-1)}{x^3} - \frac{6(5x-3)}{x^3} + \frac{6(x-1)}{x^3} = \frac{24x-12-30x+18+6x-6}{x^3} = 0.\ \checkmark \]

Step 4: Third reduction: $w=x^{-3}s$

With $w'=-3x^{-4}s+x^{-3}s'$ and $w''=12x^{-5}s-6x^{-4}s'+x^{-3}s''$, substitute into the $w$-equation and multiply through by $x^3$. The coefficient of $s$ reproduces Step 3's identity and vanishes; the coefficients of $s'$ and $s''$ are

\[ -6x(2x-1)+2x(5x-3) = -2x^2, \qquad x^2(2x-1). \] So the equation for $s$ is

\[ x^2(2x-1)s'' - 2x^2s' = 0 \implies (2x-1)s'' - 2s' = 0 \quad(\text{divide by }x^2). \]

Step 5: Solve for $s$

With $p=s'$: \[(2x-1)p' - 2p = 0 \implies \frac{dp}{p} = \frac{2\,dx}{2x-1} \implies p = A(2x-1).\]

\[ s' = A(2x-1) \implies s = A(x^2-x) + B. \]

Step 6: Back-substitute to $y$

\[ w = x^{-3}s = A(x^{-1}-x^{-2}) + Bx^{-3}, \qquad u = \int w\,dx = A(\ln x + x^{-1}) - \frac{B}{2}x^{-2} + C, \]

\[ y = xu = A\bigl(x\ln x + 1\bigr) + Cx - \frac{B}{2}\cdot\frac1x. \]

The lone constant term produced by integrating $Ax^{-1}$ cannot be absorbed into the $Cx$ or $B/x$ pieces — it stays attached to the $x\ln x$ term. So the third independent solution is $y_3 = x\ln x+1$, not $x\ln x$ alone (Step 7 confirms this). Renaming $C_1=C$, $C_2=-B/2$, $C_3=A$:

$y = C_1\,x + C_2\,\frac1x + C_3\bigl(x\ln x+1\bigr)$.

Step 7: Verify $y_3=x\ln x+1$

\[ y_3' = 1+\ln x, \qquad y_3'' = \frac1x, \qquad y_3''' = -\frac{1}{x^2}. \]

Substituting into $x^2(2x-1)y'''+(4x-3)xy''-2xy'+2y$: $-(2x-1) + (4x-3) - 2x(1+\ln x) + 2(x\ln x+1)$, $= (-2x+4x-2x) + (1-3+2) + (-2x\ln x+2x\ln x) = 0$. $\checkmark$

(Dropping the $+1$ leaves a residual of $-2$: the constant term is exactly what the $2y$ coefficient needs to cancel it.)

Step 8: Linear independence

At $x=1$: $(y_1,y_1',y_1'')=(1,1,0)$, $(y_2,y_2',y_2'')=(1,-1,2)$, $(y_3,y_3',y_3'')=(1,1,1)$, so

\[ W = \begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & 1\\ 0 & 2 & 1\end{vmatrix} = 1(-1-2) - 1(1-0) + 1(2-0) = -3-1+2 = -2 \neq 0, \]

confirming that $y_1=x$, $y_2=1/x$, $y_3=x\ln x+1$ form a fundamental system.