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

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Answer 1 · 2017-02-03T18:24:46

# y = x^3 + Ax^2 #

First write the DE in standard form:

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

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

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

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

# I = e^(int P(x) dx)#
# \ \ = e^(int \ -2/x \ dx)#
# \ \ = e^(-2lnx) #
# \ \ = e^ln(-1/x^2) #
# \ \ = -1/x^2 #

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

# -1/x^2y'+2y/x^3 = -1 #
# d/d(-y/x^2) = -1 #

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

# -y/x^2 = int -1 dx #

Which we can easily integrate to get:

# -y/x^2 = -x + c #
# :. y = x^3 + Ax^2 #

We can check our solution;

# y = x^3 + Ax^2 => y' = 3x^2+2Ax#

And so;

# xy'-2y= x(3x^2+2Ax) - 2(x^3 + Ax^2) #
# " " = 3x^3+2Ax^2 - 2x^3 -2Ax^2 #
# " " = x^3 \ \ # QED