f1 (x) = ex, f2 (x) = e-x, f3 (x) = sinh (x) g (x) = c1f1 (x) + c2f2 (x) + c3f3 (x) Solve for c1, c2, and c3 so that g (x) = 0 on the interval (-[infinity], [infinity]). If a nontrivial solution exists, state it. (If only the trivial solution exists, enter the trivial solution {0, 0, 0}.)

(C1, C2, C3) = (K, K, - 2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1 (x) = e^x

f2 (x) = e^ (-x)

f3 (x) = sinh (x)

g (x) = 0

We want to solve for C1, C2 and C3, such that

C1f1 (x) + C2f2 (x) + C3f3 (x) = g (x)

That is

C1e^x + C2e^ (-x) + C3sinh (x) = 0

The hyperbolic sine of x, sinh (x), can be written in its exponential form as

sinh (x) = (1/2) (e^x + e^ (-x))

So, we can rewrite

C1e^x + C2e^ (-x) + C3sinh (x) = 0

as

C1e^x + C2e^ (-x) + C3 (1/2) (e^x + e^ (-x)) = 0

So we have

(C1 + (1/2) C3) e^x + (C2 + (1/2) C3) e^ (-x) = 0

We know that

e^x ≠ 0, and e^ (-x) ≠ 0

So we must have

(C1 + (1/2) C3) = 0 ... (1)

and

(C2 + (1/2) C3) = 0 ... (2)

From (1)

2C1 + C3 = 0

=> C3 = - 2C1 ... (3)

From (2)

2C2 + C3 = 0

=> C3 = - 2C2 ... (4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = - 2K

(C1, C2, C3) = (K, K, - 2K)

For K in the interval (-∞, ∞)