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

Find the general solution of $x(x-1)\,y'' - (2x-1)\,y' + 2y = 2x^3-3x^2$, given that $y_1=x^2$ is a particular solution of the corresponding homogeneous equation.

Step 1: Verify $y_1$

With $y_1=x^2$, $y_1'=2x$, $y_1''=2$: $x(x-1)(2) - (2x-1)(2x) + 2x^2 = 2x^2-2x-4x^2+2x+2x^2 = 0.$ $\checkmark$

Step 2: Reduction of order

Set $y=x^2v$, so $y'=2xv+x^2v'$, $y''=2v+4xv'+x^2v''$. Substituting into the homogeneous equation, the $v$ terms cancel (they reproduce Step 1's identity), leaving: \[x^3(x-1)\,v'' + x^2(2x-3)\,v' = 0 \implies x(x-1)\,v'' + (2x-3)\,v' = 0.\]

Step 3: Solve for $w=v'$

Since $\dfrac{2x-3}{x(x-1)}=\dfrac3x-\dfrac1{x-1}$: \[\ln|w| = -3\ln|x|+\ln|x-1|+\mathrm{const} \implies w = A\,\frac{x-1}{x^3}.\]

Step 4: Integrate for $v$, find $y_2$

\[ v' = A\left(\frac1{x^2}-\frac1{x^3}\right) \implies v = A\left(-\frac1x+\frac1{2x^2}\right)+B. \]

\[ y_2 = x^2v = A\left(\frac12-x\right)+Bx^2. \]

Up to a constant multiple, the second homogeneous solution is $y_2=2x-1$. Verify: with $y_2'=2$, $y_2''=0$, $0-(2x-1)(2)+2(2x-1) = (2x-1)(-2+2) = 0.$ $\checkmark$

Step 5: Homogeneous general solution

\[ Y_h = C_1x^2 + C_2(2x-1), \qquad W(x^2,2x-1) = 2x^2-2x(2x-1) = 2x(1-x) \neq 0\ \ (x\neq0,1). \]

Step 6: Particular solution by matching coefficients

Since the right-hand side $2x^3-3x^2$ is a cubic polynomial and the leading coefficient $x(x-1)$ is quadratic, try $y^*=ax^3+bx^2+cx+d$, so $y^{*\prime}=3ax^2+2bx+c$, $y^{*\prime\prime}=6ax+2b$. Substituting into the full equation and collecting powers of $x$:

\[ x^3:\ 2a=2,\qquad x^2:\ -3a=-3,\qquad x^1:\ 0=0,\qquad x^0:\ c+2d=0. \] So $a=1$, with $b$ free and $c=-2d$; the simplest choice $b=c=d=0$ gives $y^*=x^3$.

Step 7: Verify $y^*$

With $y^{*\prime}=3x^2$, $y^{*\prime\prime}=6x$: $x(x-1)(6x) - (2x-1)(3x^2) + 2x^3 = 6x^3-6x^2-6x^3+3x^2+2x^3 = 2x^3-3x^2.$ $\checkmark$

Step 8: General solution

The solution is recorded in the result below.