Show that \(\displaystyle y = x(\ln(x))^{2} + Cx\) is a general solution to
\(\displaystyle xy' - y = 2x\ln(x)\)
\(\displaystyle y' - \dfrac{y}{x} = \dfrac{2x\ln(x)}{x}\)
\(\displaystyle e^{\int -\dfrac{1}{x} (dx)} = e^{-\ln(x)} = e^{\ln(x^{-1})} = x^{-1} = \dfrac{1}{x}\)
\(\displaystyle y'\dfrac{1}{x} - \dfrac{y}{x}\dfrac{1}{x} = \dfrac{2x\ln(x)}{x}\dfrac{1}{x}\)
\(\displaystyle \dfrac{d}{dx}[\dfrac{1}{x}y ] = \dfrac{2x\ln(x)}{x^{2}}\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2x\ln(x)}{x^{2}} (dx)\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int 2\ln(x)(x^{-1}) (dx)\)
Using integration by parts
\(\displaystyle u = 2 \ln x\)
\(\displaystyle du = \dfrac{2}{x}\)
\(\displaystyle dv = x^{-1}\)
\(\displaystyle v = \ln x\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = 2\ln(x)\ln(x) - \int \ln(x)\dfrac{2}{x}\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = 2\ln(x)\ln(x) - \int 2\ln(x)\dfrac{1}{x}\)
\(\displaystyle xy = 2\ln(x)\ln(x) - 2\ln(x) + C\) ???
\(\displaystyle xy' - y = 2x\ln(x)\)
\(\displaystyle y' - \dfrac{y}{x} = \dfrac{2x\ln(x)}{x}\)
\(\displaystyle e^{\int -\dfrac{1}{x} (dx)} = e^{-\ln(x)} = e^{\ln(x^{-1})} = x^{-1} = \dfrac{1}{x}\)
\(\displaystyle y'\dfrac{1}{x} - \dfrac{y}{x}\dfrac{1}{x} = \dfrac{2x\ln(x)}{x}\dfrac{1}{x}\)
\(\displaystyle \dfrac{d}{dx}[\dfrac{1}{x}y ] = \dfrac{2x\ln(x)}{x^{2}}\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2x\ln(x)}{x^{2}} (dx)\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int 2\ln(x)(x^{-1}) (dx)\)
Using integration by parts
\(\displaystyle u = 2 \ln x\)
\(\displaystyle du = \dfrac{2}{x}\)
\(\displaystyle dv = x^{-1}\)
\(\displaystyle v = \ln x\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = 2\ln(x)\ln(x) - \int \ln(x)\dfrac{2}{x}\)
\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = 2\ln(x)\ln(x) - \int 2\ln(x)\dfrac{1}{x}\)
\(\displaystyle xy = 2\ln(x)\ln(x) - 2\ln(x) + C\) ???