(I'm going to translate y' to dy/dx as it makes it easier to read for me, you could change it back if you wanted.)
[tex]x \frac{dy}{dx} -y=0[/tex]
[tex] \frac{dy}{dx} = \frac{y}{x} [/tex]
[tex]\int \frac{1}{y} dy= \int \frac{1}{x} dx[/tex] (separate the variables)
[tex]\ln y=\ln x +c[/tex]
[tex]\ln y = \ln kx[/tex] (by letting c = ln k and using log laws)
[tex]y=kx[/tex] (raise everything to power e)
[tex]3=k\times1 \implies k=3[/tex] (applying boundary conditions)
Particular solution: [tex]y=3x[/tex]
