1
$\begingroup$

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?

  • 0
    For one thing, those two functions $a$re not solutions of that differential equation.2012-09-24

1 Answers 1

1

The flaw is that you are indeed mistaken, in that the equation you present is

  • second order, not first order (this is not relevant, but worth pointing out)
  • not linear.

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 $b\not=0$.

Another immediate way of observing your equation is not linear is observing that the zero function is not a solution.

  • 0
    @Mel: if I've understood you correctly, then yes.2012-09-24