# Solutions of Linear Homogenous Differential Equations as a Vector Space

If I'm not mistaken, the set of all functions $f(x)$ satisfying the first order homogeneous ODE:

$$f''(x) - 2x = 0$$

is a Vector Space (as in, the elements of the Vector Space are its solutions).

Two solutions for the above ODE are $f(x) = x^2 + 7$ and $f(x) = x^2 + 9$.

Therefore, if they are elements of the Vector Space, a linear combination of them say: $2x^2 + 16$, should also be a solution to the ODE above. However, it is not.

Where is the flaw above?

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This equation is not homogeneous because of the $2x$. An homogenous equation should only involve $f$ and its derivatives. The homogenous equation associated to this one is simply $f" = 0$ (and you can check that the difference of your two solution is a solution of the homogeneous equation). – Joel Cohen Sep 24 '12 at 18:43
@Joel Thanks for clearing that up. So if the equation were, say: f''(x) - 2 f(x) == 0 then the above should hold true? – user64219 Sep 24 '12 at 18:49
Yes, exactly, and you can check $f(x) = e^{t\sqrt{2}}$ and $f(x) = e^{-t\sqrt{2}}$ are solution as well as all their linear combinations. – Joel Cohen Sep 24 '12 at 18:51
For one thing, those two functions are not solutions of that differential equation. – Thomas Andrews Sep 24 '12 at 19:34

In fact, your comment amounts to a proof that the equation in question is not linear, because the sum of solutions need not be a solution. But you can also think in the following terms: a linear homogeneous equation is of the form $Ly=0$ for some linear operator $L$. But the operator you are applying to $y$ is the operator $f\mapsto f\prime\prime - 2\operatorname{id}$, and that subtraction of $2\operatorname{id}$ makes it non-linear, just like the operator on vectors $x\mapsto Ax - b$ is non-linear if $b\not=0$.