Linear, first order ODE, given general solution Construct a linear first order ODE of $$xy'+a_0(x)y=g(x)$$
such that the general solution is $$y=x^3+\frac{c}{x^3}$$
If I substitute the general solution into the question, 
$$xy'+a_0(x)(x^3+\frac{c}{x^3})=g(x)$$
this form looks unusual to solve and a(x), g(x) are not given in any expression.  how can I to approach this?
 A: *

*First you should compute $y'$ and plug it in. 

*Since the general solution is $y = x^3 + c x^{-3}$, consider two solutions $y_1 = x^3 + c_1 x^{-3}$ and $y_2 = x^3 + c_2 x^{-3}$. Linearity tells you that $y_1 - y_2 = (c_1 - c_2) x^{-3}$ is a solution to the corresponding homogeneous problem. This should allow you to find $a(x)$. 

*Using the knowledge from $a(x)$ you can then find the inhomogeneous term $g(x)$ using the general form of the solution. 

A: Isolating the constant gives
$$
C=x^3y-x^6.
$$
Differentiating to get rid of the constant results in
$$
0=x^3y'+3x^2y-6x^5
$$
The canonical first order linear equation form of that is then
$$
y'+\frac3xy=6x^2
$$
A: If $y = x^3 + cx^{-3}$ then
$$
    xy' = x\left(3x^2 -3cx^{-4}\right) = 3x^3 - 3cx^{-3}
$$
So
$$
   xy' + a_0(x) y 
   = 3x^3 - 3cx^{-3}+ a_0(x) \left(x^3 + cx^{-3}\right)
   = (a_0(x)+3)x^3 + c(a_0(x)-3)x^{-3}
$$
Since this is to be equal to a function $g(x)$ independent of $c$, we must have $a_0(x)=3$.  It follows that $g(x)=6x^3$ works.
