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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
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1 Answer 1

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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.

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Thanks for that. I actually don't know how I didn't realise my equation wasn't first order, how embarrassing! For closure's sake, would the above properties hold in a equation where: The derivative of f(x) is at most 2, and the only other terms in the equation are constant multiples of f(x) and f'(x) –  user64219 Sep 24 '12 at 19:08
    
@Mel: if I've understood you correctly, then yes. –  Ben Millwood Sep 24 '12 at 21:43
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