Calculus help?

Question

Let f and g be functions that are differentiable for all real numbers x and that have the following properties -
(i) f'(x) = f(x) - g(x)
(ii) g'(x) = g(x) - f(x)
(iii) f(0) = 5
(iv) g(0) = 1

(a) Prove that f(x) + g(x) = 6 for all x.
(b) Find f(x) and g(x). Show your work.

Answer 1

f'(x)=f(x)-g(x)

g'(x)=g(x)-f(x) --> g(x)=g'(x)+f(x)

f''(x)=f(x)-g'(x)-f(x)=g'(x)

integrating f(x)=-g(x)+C

f(0)=5, g(0)=6 so 5 = -1 + C

C = 6

f(x) + g(x) = 6 for all x

Answer 2

f'(x)=f(x)-g(x) ===> f(x)=f'(x)+g(x) ==> f(0) ==> 5=0+g(x) ==> g(x)=5

the same for g(x) ==> g(0) ==> 1=f(x) + 0 ==> f(x)=1

f(x)+g(x)=1+5=6 !!

easy ! just imagine ! you can do it .

Answer 3

(a)

add (i) and (ii):
f'(x) + g'(x) = 0 => f(x)+g(x) = const = f(0) + g(0).

(b)
subtract (ii) from (i)

f'(x) - g'(x) = 2(f(x)-g(x))
let q(x) = f(x)-g(x), then
q'(x) = 2q(x),
dq/dx = 2q
dq/q = 2 dx
ln q = 2x + C
q = A exp(2x),

A = q(0) = f(0) - g(0) = 4

Now you need to solve a system of ordinary linear equations:
f + g = 6
f - g = 4exp(2x)


f(x) = 3 + 2 exp(2x)
g(x) = 3 - 2 exp(2x)

Answer 4

(i) f'(x) = f(x) - g(x)
(ii) g'(x) = g(x) - f(x)
(iii) f(0) = 5
(iv) g(0) = 1

The answer is 37
....