Suppose that the function has a second derivative and that?

Question

f ''(x) + f(x) = e^(-x) for all x
f(0) = 0 and f '(0) = 2

Find the fourth Taylor polynomial at x=0.

Please help me on this.
Thanks

Answer 1

The question is much easier than having to solve the Diff EQ (since they ask for the Taylor Polynomial).

The fourth order Taylor polynomial at x = 0 is:

f(0) + f'(0)x + (f''(0)/2)x^2 + (f'''(0)/6)x^3 + (f''''(0)/24)x^4.

Now we know that f(0) = 0 and f'(0) = 2.

Now to get the others, use the differential equation.

f''(0) + f(0) = e^(-0), so f''(0) + 0 = 1, so f''(0) = 1.

Now taking the derivative, you get f'''(0) + f'(0) = -e^{-x} and so

f'''(0) + 2 = -1, so f'''(0) = -3.

Finally, taking two derivatives, you get f''''(0) + f''(0) = e^{-x}, so f''''(0) + 1 = 1.

So f''''(0) = 0.

Putting it all together, you get the fourth-degree Taylor polynomial to be:

2x + (-3/2)x^2 + (1/6)x^3. (since the constant and the x^4 terms are multiplied by 0)

Answer 2

There are two parts to finding the solution to the Diff EQ. The general solution and the specific solution (or whatever the appropriate term is these days).

To get the general solution, solve D² + D = 0, which implies sine and cosine:
Asin(x) + Bcos(x)
for the specific solution, we use
ce^(-x) so that
ce^(-x) + ce^(-x) = e^(-x) which means that c=1/2
Therefore, we have that

f(x) = Asin x + Bcos x + (e^(-x))/2
f'(x) = Acos x - Bsin x - (e^(-x))/2

Since f(0) = 0
B + 1/2 = 0 or B = -1/2
from f'(0) = 2 we have:
A - 1/2 = 2 or A = 5/2

Thus,
f(x) = (5sin(x) - cos(x) + e^(-x))/2