# Prove the identity f"(x)-f'(x)-2f(x)=0 f(x)=2e^2x-3e^-x

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### 1 Answer

To prove the given identity, first we need to determine the terms in identity, namely f"(x) and f'(x).

We'll begin with f'(x):

f'(x) = (2e^2x-3e^-x)'

f'(x) = 4e^2x + 3e^-x

f"(x) = [f'(x)]'

f"(x) = (4e^2x + 3e^-x)

f"(x) = 8e^2x - 3e^-x

Now, we'll substitute the expressions of f"(x) andf'(x) into the identity that has to be verified:

8e^2x - 3e^-x - 4e^2x - 3e^-x - 2(2e^2x-3e^-x)

We'll remove the brackets and we'll combine like terms:

4e^2x - 6e^-x - 4e^2x + 6e^-x

We'll eliminate matching terms:

4e^2x - 6e^-x - 4e^2x + 6e^-x = 0

**4e^2x - 6e^-x - 4e^2x + 6e^-x = 0 <=> f"(x) - f'(x) - 2f(x) = 0 q.e.d.**