How to prove linearity? Given a third-order differential equation, of the form $y''' + f(t,y,y',y'') = 0.$ that admits solution:
$$Y(t) = y(t) + C_1 f_1(t) + C_2 f_2(t) + C_3 f_3(t)$$
where $y(t)$ is a particular solution, and  $f_1(t), f_2(t), f_3(t)$ are linearly independent functions, then
How to prove that the differential equation is linear?,
that is I want to prove that $$f(t,y,y',y'') = A(t) y'' + B(t) y' + C(t) y + D(t)$$
 A: Write
$$\begin{cases}Y(t)- y(t) = C_1 f_1(t) + C_2 f_2(t) + C_3 f_3(t)\\
Y'(t)- y'(t) = C_1 f_1'(t) + C_2 f_2'(t) + C_3 f_3'(t)\\
Y''(t)- y''(t) = C_1 f_1''(t) + C_2 f_2''(t) + C_3 f_3''(t)\\
Y'''(t)- y'''(t) = C_1 f_1'''(t) + C_2 f_2'''(t) + C_3 f_3'''(t).\end{cases}$$
Now you can express that this system has nontrivial solutions for $C$ by the condition
$$\Delta=0$$ which can be developed as
$$(Y(t)- y(t))M_0(t)+(Y'(t)- y'(t))M_1(t)+(Y''(t)- y''(t))M_2(t)+(Y'''(t)- y'''(t))M_2(t)=0$$ (where $M_k$ are minors built on the RHS), a linear ODE.
A: Take the differential equation :
$$(y^{(3)})^2=0 $$
Then the solutions to this equation are exactly the polynomials of degree $\leq 2$. Then set $f_1(t):=1$, $f_2(t):=t$ and $f_3(t):=t^2$. And $y(t)=0$ is a solution. Hence all the solutions of my equation are of the form :
$$Y(t)=y(t)+C_1f_1(t)+C_2f_2(t)+C_3f_3(t) $$
With $f_1$, $f_2$ and $f_3$ being linearly independant.
However I wouldn't say that the differential equation is linear (because of the square...). Nevertheless the set of solution is a vector space. 
A: Assume you have two solutions, prove that their sum is a solution, proceed with the other requirements of linearity. 
