Question

determine the non-zero constants a, b, c such that
\[ae^x+bcosh(x)+csinh(x)=0\]
I don't know how to do

Answer 1

\[ae^x+b(\frac{e^x+e^{-x}}{2})+c(\frac{e^x-e^{-x}}{2})=0\]
\[ae^x+\frac{be^x}{2}+\frac{be^{-x}}{2}+\frac{ce^x}{2}+\frac{ce^{-x}}{2}=0\]

Answer 2

so, \[e^x(2a+b+c)+e^{-x}(b+c)=0\]and ....stuck

Answer 3

Haven't looked much into this yet, but maybe it helps to switch it over to the imaginary part,

\[\large \cosh(ix)=\cos(x) \]
and \[\Large \sinh(ix)=i\sin(x) \]
then \[ \Large e^{ix}=\cos(x)+i\sin(x)\]

Answer 4

but it's real part, we don't have i there.

Answer 5

true but for tricky question such as this it's often worth a thought to switch the domain from \(\mathbb{R}\) to \(\mathbb{C}\)

Answer 6

I don't know how to convert to C, please show me

Answer 7

I just tried and I get stuck, just in a different domain :-) didn't help much.

Answer 8

:(

Answer 9

for this thing to be 0 for all "x" , the terms have to cancel
try
b = c
a = -b

set "b" to any arbitrary number

Answer 10

I think e^x and e^(-x) is never =0, therefore 2a +b +c = 0 and b+c =0
--> a =0 and b = -c
does it make sense?

Answer 11

@Loser66 , You're right that \(e^x \not=0 \) and \(e^{-x} \not = 0 \) but if you set \(a=0\) then that contradicts with the statement given by the \[\Large a,b,c \in \mathbb{R} / 0 \]

Answer 12

@Loser66, i dont think you understood my post
cancel terms
\[(2a+b+c)e^{x} + (b-c)e^{-x} = 0\]
b-c = 0
2a+b+c = 0

c = b
a = -b

let b =-1, then a=1, c=-1
that is 1 solution, there are infinite solutions for (a,b,c)
(a,b,c) = (-n,n,n) for any real num "n"

Answer 13

http://www.wolframalpha.com/input/?i=-n*e%5E%28x%29+%2Bn*cosh%28x%29%2Bn*sinh%28x%29+

Answer 14

I got you. thanks @dumbcow. I forgot the condition of a, b, c \(\neq\) 0. :-)))

Answer 15

1 equation, 3 unknowns, infinite number of solutions -- @dumbcow 's got it right.