What is the solution of y'=(y+9x)^2, v=y+9x...Please

y'=(y+9x)^2, v=y+9x.
v=y+9x.
Hence dv/dx=dy/dx+9
i.e dy/dx=dv/dx−9
hence
dv/dx−9=(v)²
→dv/dx=(v)² +9
or dv/(9+v²)=dx
(1/3)(tan‾¹(v/3) =x+C
or (1/3)(tan‾¹{(y+9x)/3} =x+C
or tan(3x+3C) =(y+9x)/3
or y= 3tan(3x+K) −9x where K= 3C another constant

v ' - 9 = v^2 --->
dx = dv/(v^2+9)
integrate both sides
x+C = (1/3)tan^-1 (v/3) , then substitute y+9x back