So, I have something like this:
$$y = \frac{Cx^2 + Cx}{Cx^2+ 1}$$
general solution.
Can I find the differential equation itself from this general solution?
If yeah – how?
Find differential equation from general solution
ordinary differential equations
Best Answer
Remember that an expression with $n$ arbitrary constants will yield a differential equation of order $n$. So to get the $n^{th}$ order derivative you'll have to differentiate the expression $n$ times, and in that process you'll obtain $n$ more relations so that now you have a total of $n+1$ relations from which you can eliminate the $n$ arbitrary constants to obtain the differential equation.
In your example
$$y = \frac{Cx^2 + Cx}{Cx^2+ 1}$$
there is only one constant $C$. Therefore, we can rearrange to form
$$y=C(x^2+x−x^2y)$$
by which we differentiate once to get
$$y'=C(2x+1-2xy-x^2y')$$
and since
$$C=\frac{y}{x^2+x−x^2y}$$
the above equation is
$$y'=\frac{2xy+y-2xy^2-x^2yy'}{x^2+x-x^2y}$$
which simplifies to the differential equation
$$y'=\frac{2xy+y-2xy^2}{x^2+x}$$