If the 2 meant *2 then:
Expand and move y to left side to get
y’-2*y=2*x-6.
The homog eqn is yh’-2*yh=0 so yh=k1*exp(2*x) by trying y=exp(m*x) or separating.
Assume yp=a*x+b so yp’=a then
a-2*(a*x+b)=2*x-6 or
-2*a*x+a-2*b=2*x-6 so
-2*a=2 so a=-1 and a-2*b=-6 so
-1–2*b=-6 so -2*b=-5 and b=5/2 so we have yp=-x+5/2 which yields the general soln y=yh+yp=k1*exp(2*x)-x+5/2.
For y(0)=0, we see k1+5/2=0 so k1=-5/2 and the solution is
y=5*(1-exp(2*x))/2-x.
This heads exponentially to minf for larger x.
If the 2 is ^2 then
y’=(x+y-3)^2 and let y=v-x+3 so y’=v’-1 and y’=(x+y-3)^2 becomes v’-1=v^2 or
v’=1+v^2 so separate as dv/(1+v^2)=dx and integrate to get
atan(v)=x+k2 so v=tan(x+k2)=y+x-3 so y=tan(x+k2)-x+3 and y(0)=0 becomes
0=tan(k2)+3 and tan(k2)=-3 so k2=-atan(3) which makes y=tan(x-atan(3))-x+3.
This has singularities for x=atan(3)+%pi*(2*n+1)/2 for integer