Let $L(y) = x^2 y'' - 3xy' + 4y$
$y_1(x) = x^2, y_1'(x) = 2x, y_1''(x)= 2$
Sub it into $L(y)$, then $L(y) = 2x^2 - 3x(2x) + 4x^2 = 0$, hence $y_1(x) = x^2$ satisfies the homo equation.
$y_2(x) = x^2 ln(x), y_2'(x) = x + 2xln(x), y_2''(x) = 3 + 2ln(x)$
Sub it into $L(y)$, then $L(y) = x^2(3 + 2ln(x)) - 3x(x + 2xln(x) + 4x^2 ln(x) = 0$, hence $y_2(x) = x^2 ln(x)$ satisfies the homo equation.
$W[y_1, y_2](x) = \left|\begin{matrix}y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x)\end{matrix}\right|= \left|\begin{matrix}x^2 & x^2 ln(x) \\ 2x & x + 2xln(x)\end{matrix}\right| = x^3$
Let the particular solution $y_p(x) = u_1 y_1(x) + u_2 y_2(x)$,
Since $x^2 y'' - 3xy' + 4y = x^2 ln(x)$, thus $y'' - \frac{3}{x}y' + \frac{4}{x^2}y = ln(x)$
$p(x) = - \frac{3}{x}, q(x) = \frac{4}{x^2}, g(x) = ln(x)$
$u_1 = -\int{\frac{y_2(x)g(x)}{W[y_1, y_2](x)}}dx = - \int{\frac{x^2 ln(x)ln(x)}{x^3}}dx = - \frac{1}{3}(ln(x))^3 +c_1$
$u_2 = \int{\frac{y_1(x)g(x)}{W[y_1, y_2](x)}}dx = \int{\frac{x^2 ln(x)}{x^3}}dx = \frac{(ln(x)^2)}{2} +c_2$
$y(x) = (- \frac{1}{3}(ln(x))^3 +c_1) x^2 + (\frac{(ln(x)^2)}{2}+c_2)x^2 ln(x) = \frac{1}{6}x^2(ln(x))^3 + c_1 x^2 + c_2 x^2 ln(x) $
The particular solution is thus
$y_p(x) = \frac{1}{6}x^2(ln(x))^3$
