Answer:
(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 (-∞, ∞)
