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Prove that every solution $f$ to the Ordinary Differential Equation $f''-f=0$ can be written as a linear combination of the solutions $f_1(x)=e^x$ and $f_2(x)=e^{-x} $ .i.e. if $f$ is any solution then there exist two real numbers $c_1$,$c_2$ such that $f=c_1f_1+c_2f_2$

We are given a hint which says if $f$ is a solution of $f''-f=0$, what can you say about the functions $e^{-x}(f+f')$ and $e^x(f-f')$.

I'm not sure where to proceed. Substituting in the values of $f$ and $f'$ into the the equations given in the hint gives me $e^{-x}(f+f')=2c_1$ and $e^x(f-f')=2c_2$ and differentiating the statements just gives me a $0$

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    $\begingroup$ It's trivial to show that $f_1,f_2$ solve the DE. You could likely use the Wronskian to show they're linearly independent. Because the DE is second-order, the solution space must have dimension $2.$ Would that be sufficient? $\endgroup$ Oct 22 '19 at 18:41
  • $\begingroup$ @AdrianKeister how does showing that they are linearly independent relate to the proof? (we are asked to show that they are linearly independent earlier in the question) $\endgroup$ Oct 22 '19 at 19:12
  • $\begingroup$ A linearly independent set of the right dimension is a basis, and if you have a basis, then the claim is proved; every vector in the space can be written in terms of the basis, by definition. $\endgroup$ Oct 22 '19 at 19:15
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$f''-f=0$ means $f''=f$

define $g^-(x) = e^{-x}(f+f')$ and $g^+(x) = e^x(f-f')$.

We can find $$ g^-(x)' = -e^{-x}(f+f') +e^{-x}(f'+f'') = -e^{-x}(f+f') +e^{-x}(f'+f) = 0 $$

Thus, $g^-(x) = C^-$ is a constant. $\quad==>\quad f+f' = C^-e^x$

Similarly, $g^+(x) = C^+$ is also a constant .$\quad==>\quad f-f' = C^+e^-x$

Adding the 2 equations above, we get $$ f = \frac{1}{2}\left(C^-e^x + C^+e^-x\right) $$

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Here is a sketch.

Let $g=f'-f$ so that $g'+g=0$. Then if $h=e^{x}g$ we have $h'=e^{x}g'+e^{x}g=0$ whence $h=A$ (constant) and $g=Ae^{-x}$.

Let $k=f'+f$ so that $k'-k=0$ and in a similar way $k'=Be^x$

Then $2f=k-g$

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