What is the general solution of the differential equation # y'-2xy=x^3 #?

We have:

# y'-2xy=x^3 # ..... [1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

We can readily generate an integrating factor when we have an equation of this form, given by;

# I = e^(int P(x) dx) #
# \ \ = exp(int \ -2x \ dx) #
# \ \ = exp( -x^2 ) #
# \ \ = e^( -x^2 ) #

And if we multiply the DE [1] by this Integrating Factor, #I#, we will have a perfect product differential;

# e^( -x^2 )y'-2xe^( -x^2 )y=x^3e^( -x^2 ) #
# :. d/dx(e^( -x^2 )y) = x^3e^( -x^2 ) #

We can now integrate to get:

# e^( -x^2 )y = int \ x^3e^( -x^2 ) \ dx + C #

The RHS integral can be evaluated by an application of Integration By Parts (omitted) which gives us:

# int \ x^3e^( -x^2 ) \ dx = -1/2(x^2+1)e^(-x^2) #

So we have:

# e^( -x^2 )y = -1/2(x^2+1)e^(-x^2) + C #

Using the initial condition #y(0)=2# we can evaluate the constant #C#

# 2e^0 = -1/2(0+1)e^0 + C => C = 5/2#

Leading to the Particular Solution:

# e^( -x^2 )y = -1/2(x^2+1)e^(-x^2) +5/2 #

# :. y = -1/2(x^2+1)e^(-x^2) +5/(2e^( -x^2 )) #
# \ \ \ \ = -1/2(x^2+1) +5/2e^( x^2 ) #
# \ \ \ \ = 5/2e^( x^2 )-1/2x^2-1/2 #