# B Short check on solved ODE problem

#### 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!

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#### Mark44

Mentor
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.
Silviu said:
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!

"Short check on solved ODE problem"

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