The general solution to the ODE t y ' +2y = 4(t^2)  is  y(t) = (t^2)+((t^2)/C).  These solution curves, for various values of C, are graphed here.

plot [K_]:=ContourPlot[y==t^2+K/(t^2),{t,-2,2},{y,-3,3},Axes->True,AxesOrigin->{0,0},Frame->False,ContourStyle->{Thick,Blue}]
rpoint[x_,y_]:= Graphics[{Red,PointSize[.03],Point[{x,y}]}];
tangent[x_,y_]:= ParametricPlot[{x,y+t},{t,-.5,.5},PlotStyle->{Orange,Thick}]
domain[m_,M_]:= Plot[0,{x,m,M},PlotStyle->{Red,Thickness[.02]}];
Show[plot[0],plot[1],plot[-1],plot[.5],plot[-.5],plot[.25],plot[-.25]]

Solve the IVP t y ' +2y = 4(t^2)  according to the following initial values:  (a)  y(-1) = 1, (b)  y(-1) = 2, (c)  y(0) = 0, and (d)  y(0) = 1.  Remember, you must specify the  domains for any particular solution.

rpoint[x_,y_,color_]:= Graphics[{color,PointSize[.03],Point[{x,y}]}];
	
	initialValueA = rpoint[-1,1, Red];
	initialValueB = rpoint[-1,2,Black];
	initialValueC = rpoint[0,0,Green];
	initialValueD = rpoint[0,1,Orange];

Show[plot[0],plot[1],plot[-1],plot[.5],plot[-.5],plot[.25],plot[-.25],initialValueA,initialValueB,initialValueC,initialValueD]