Find an initial condition for the ODE $xy'+3y=6x^3$ Find an initial condition of the form 
$$y(x_0)=y_0$$
for the ODE
$$xy'+3y=6x^3$$
so that the solution is
$$ y=x^3-\frac{1}{x^3}$$
I found that the general solution to the ODE is
$$y=x^3+\frac{C}{x^3}$$
So, how should I proceed to find $C$?
 A: Since $y(x_0)=y_0$ and $y=x^3+\frac{C}{x^3}$, plugging the initial condition yields to
$$y_0=y(x_0)=x_0^3+\frac{C}{x_0^3}.$$
From this,
$$C=(y_0-x_0^3)\;x_0^3.$$
Then, the solution of your ODE is
$$y=x^3+\frac{(y_0-x_0^3)\;x_0^3}{x^3}.$$
A: You have found the general solution. That is great!
Now you want to find the initial condition such that the constant, C, in the solution = -1.
Your general solution has the form; y(x) = $x^3$ + $\frac{C}{x^3}$
If we consider the form of the initial condition equated as the form of the general solution; 
y($x_0$) = $x_0^3$ + $\frac{C}{x_0^3}$ = $y_0$
We want to find $x_0$ and $y_0$ such that the above equation solves C = -1, by inspection (not always easy) we may choose $y_0$ = 0 and $x_0$ = 1, such that:
y($x_0$ = 1) = 1 + C = 0 = $y_0$ $=>$ C = -1
Hence giving your solution the form:
y(x) = $x^3$ - $\frac{1}{x^3}$
In this particular problem we were asked to solve one equation in one unknown, if we had more than one constant we would have to find additional initial conditions, such as an initial condition on some boundary or on the derivative of the solution.
A: You found that the given function is actually an element of the solution family. Now you have to just pick some $x_0$ and insert into the given function.
\begin{array}{c|c}
x_0&y_0=y(x_0)=x_0^3-\frac1{x_0^3}\\\hline
1&0\\
2&8-\frac18=7+\frac78=\frac{63}{8}\\
\frac12&\frac18-8=-\frac{63}{8}
\end{array}
etc.
