Answer
See answer below
Work Step by Step
Take the derivatives of the function.
$$y(x)=c_1x^2+c_2x^{2}\ln (x)+\frac{1}{6}x^2(\ln(x))^3$$
$$y'(x)=\frac{1}{6}x(12c_2\ln(x)+12c_1+6c_2+2 \ln^3(x)+3\ln^2(x))$$
$$y''(x)=2c_2\ln (x)+3c_2+2c_1+\frac{1}{3}\ln^3(x)+\frac{3}{2}\ln^2(x)+\ln (x)$$
Substitute these functions into the differential equation to get $$x^2y''+4xy=0$$
$$x^2[2c_2\ln (x)+3c_2+2c_1+\frac{1}{3}\ln^3(x)+\frac{3}{2}\ln^2(x)+\ln (x)]-3x[\frac{1}{6}x(12c_2\ln(x)+12c_1+6c_2+2 \ln^3(x)+3\ln^2(x))]+4[c_1x^2+c_2x^{2}\ln (x)+\frac{1}{6}x^2(\ln(x))^3]=x^2\ln(x)$$
$$\frac{\ln(x)x^2}{6}[2\ln^2(x)+9\ln(x)+6]-\frac{3x^2}{6}[2\ln^3(x)+3\ln^2(x)]+\frac{4x^2}{6}=x^2\ln(x)$$
$$\ln(x)x^2=x^2\ln(x)$$
$$0=0$$
This equation is always true, so the equation is a solution to the differential equation.
