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

1 Answer
Feb 3, 2017

y = x^3 + Ax^2

Explanation:

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