Find the particular solution to y′=2xy' = 2xy′=2x with y(1)=3y(1) = 3y(1)=3.
y=x2+2y = x^2 + 2y=x2+2
y=x2+3y = x^2 + 3y=x2+3
y=2x2+1y = 2x^2 + 1y=2x2+1
y=x2+1y = x^2 + 1y=x2+1