Can This ODE Solution f'(x)=f(2-x) Be Correctly Solved?

[Silviu]

Hello! I have this problem f'(x)=f(2-x) and I need to find f. This is what I did
x -> 2-x
f'(2-x)=f(2-2+x)=f(x) => f''(x)=f'(2-x)=f(x) => f''(x)=f(x) => $f(x)=c_1e^x+c_2e^{-1}$.

So, $c_1e^x-c_2e^{-x}=c_1e^{2-x}+c_2e^{x-2}$ => $-c_2e^{-x}=c_1e^2e^{-x}$ => $-c_2=c_1e^2$. And, similarly, $c_1=c_2e^{-2}$. So $-c_2=c_2e^2e^{-2}$ => $c_2=c_1=0$ => f(x)=0.

Is this correct? There is another solution? Thank you!

 

[Mark44]

Hello! I have this problem f'(x)=f(2-x) and I need to find f. This is what I did
x -> 2-x
f'(2-x)=f(2-2+x)=f(x) => f''(x)=f'(2-x)=f(x) => f''(x)=f(x) => $f(x)=c_1e^x+c_2e^{-1}$.

I see three problems.
1) You replaced x by 2 -x, but you aren't consistent with this change. If you make a substitution, use a variable with a different name.
2) From f''(x) = f(x) (which does not follow from the original diff. equation), the general solution should be $f(x)=c_1e^x+c_2e^{-x}$, which is different from what you have.
3) In your work below, you solve for the coefficients c₁ and c₂. Since there are no initial conditions given, there's no way to solve for these constants.

So, $c_1e^x-c_2e^{-x}=c_1e^{2-x}+c_2e^{x-2}$ => $-c_2e^{-x}=c_1e^2e^{-x}$ => $-c_2=c_1e^2$. And, similarly, $c_1=c_2e^{-2}$. So $-c_2=c_2e^2e^{-2}$ => $c_2=c_1=0$ => f(x)=0.

Is this correct? There is another solution? Thank you!