Integrate both sides with respect to x :
[tex]y''(x) = \ln(x) \implies \displaystyle \int y''(x)\,\mathrm dx = \int \ln(x)\,\mathrm dx[/tex]
On the right side, integrate by parts with
[tex]f = \ln(x) \implies \mathrm df = \dfrac{\mathrm dx}x \\\\ \mathrm dg = \mathrm dx \implies g = x[/tex]
[tex]\implies \displaystyle \int\ln(x)\,\mathrm dx = fg - \int g\,\mathrm df \\\\ \int\ln(x)\,\mathrm dx= x\ln(x) - \int \mathrm dx \\\\ \int\ln(x)\,\mathrm dx= x\ln(x) - x + C_1[/tex]
Then
[tex]\displaystyle y'(x) = x\ln(x) - x + C_1[/tex]
Integrate both sides with respect to x by parts again, this time with
[tex]f = \ln(x) \implies \mathrm df = \dfrac{\mathrm dx}x \\\\ \mathrm dg = x\,\mathrm dx \implies g = \dfrac{x^2}2[/tex]
[tex]\implies \displaystyle \int x\ln(x)\,\mathrm dx = fg-\int g\,\mathrm df \\\\ \int x\ln(x)\,\mathrm dx = \dfrac{x^2\ln(x)}2 - \int\frac x2\,\mathrm dx \\\\ \int x\ln(x)\,\mathrm dx = \dfrac{x^2\ln(x)}2-\dfrac{x^2}4 + C_2[/tex]
Then
[tex]y(x) = \dfrac{x^2\ln(x)}2 - \dfrac{x^2}4 - \dfrac{x^2}2 + C_1x + C_2 \\\\ \boxed{y(x) = \dfrac{(2\ln(x)-3)x^2}4 + C_1x + C_2}[/tex]
