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