# Every $f$ that is a solution of $f''-f=0$ can be written as a linear combination of the solutions $f_1(x)=e^x$ and $f_2(x)=e^{-x}$

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 $$0$$

• 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? Oct 22 '19 at 18:41
• @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) Oct 22 '19 at 19:12
• 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. Oct 22 '19 at 19:15

$$f''-f=0$$ 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 $$f = \frac{1}{2}\left(C^-e^x + C^+e^-x\right)$$

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 $$2f=k-g$$