Find the particular solution to y′=2xyy' = 2x yy′=2xy with y(0)=5y(0) = 5y(0)=5.
y=5ex2y = 5e^{x^2}y=5ex2
y=ex2+4y = e^{x^2} + 4y=ex2+4
y=5e2xy = 5e^{2x}y=5e2x
y=e2x2+5y = e^{2x^2} + 5y=e2x2+5