If f"(x)-f'(x)-2f(x)=0, f'(0)=-2, and f(0)=2, what's f(1)? (Answer: 2e^-1)

Question

anonymous · Answer

y'' - y' -2 y=0

Assumes linear equation hence  f(x)=A e^x

f(x)= A e^cx
f(0)= A e^c0 =A*1=2
A=2
f'(x)=cA e^cx

f'(0)= cA e^(c0)=-2

cA=-2

A=2
hence c=-1

anonymous · Answer

now with what you know plug in 1 for f(x)

loser66 · Answer

wonder how can you not know how to solve it.????
characteristic equation r^2 - r -2 = 0 
r =2 and r =-1
general solution is 
$$f(x)= C_1e^{-x}+C_2e^{2x}$$
$$f(0) = C_1 +C_2 = 2~~~ (*)$$
$$f'(x) = -C_1e^{-x}+2C_2e^{2x}$$
$$f'(0) = -C_1 +2C_2=-2~~~(**)$$
(*) and (**) ---> $C_1 = 2$ and    $C_2 =0$

So, the solution is $f(x) = 2e^{-x}$

replace x = -1 $f(-1) = 2e$ 

I am sorry for getting the solution which is not as yours but I don't know what my mistake is

anonymous · Answer

my solution is valid and would have gotten your answer. how can you be so ungrateful as to ignore my response.