What's f(1)?

Question

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

 

Answer: 2e^-1

 

 

Answer 1

This is a homogenous second order linear differential equation, and can be solved in a couple of ways, of which I will describe the first (most intuitive) way. Essentially, it is to make an educated "guess" at the answer, and to use this guess to arrive at the actual solution, as follows:

1. "Guess" that the solution will be proportional to e^λx for some constant λ (which is the case for this kind of equation). Substitute this for f(x), yielding
   
   λ²e^λx -  λe^λx - 2e^λx = 0
   
   
2. Factor out e^λx, yielding
   
   (λ² - λ - 2)(e^λx) = 0
   
   
3. For any finite λ, e^λx cannot equal zero, so it can be eliminated by dividing through by it.
   
   λ² - λ - 2 = 0
   
   
4. Solving this gives us λ = 2 and λ = -1, which leads to the general solution for the differential equation.
   
   f(x) = c₁e^2x + c₂e^-1x
   
5. Substitute with the given values for f'(0) and f(0).
   
   2 = c₁ + c₂
   -2 = 2c₁ - c₂
   
   
6. Solving the system gives c₁ = 0 and c₂ = 2, which gives us the particular solution:
   
   f(x) = 2e^-x
   
   
7. So f(1) = 2e^-1.

 

The second way is to use a Laplace Transform, which is actually less work, but has a little more conceptual overhead, so I won't go into describing it here. Hope this helps :)

Answer 2

First, find the characteristic polynomial for the equation and solve it:

λ²-λ-2=0   ⇒ λ = -1, 2 

Therefore, the general solution is

y = c₁ e^-x + c₂ e^2x.

Find c₁ and c₂ from the initial conditions:

y(0) = c₁ + c₂ = 2

y'(0) = -c₁ + 2c₂ = -2

which gives you c₁=2 and c₂=0.

Therefore,

y = 2 e^-x

y(1) = 2 e^-1.