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My prof. wrote this today:

$$ Ae^{\alpha_n x} + B e^{-\alpha_n x} + C' \cos(\alpha_n x) +D'\sin(\alpha_n x)$$ $$ =A'\cosh(\alpha_n x) + B'\sinh(\alpha_n x) + C' \cos(\alpha_n x) + D'\sin(\alpha_n x)$$

Can someone please help me to understand how he got the second line from the first line?

How does he arrive at $A'\cosh(\alpha_n x)+B'\sinh(\alpha_n x)$ from $Ae^{\alpha_n x}+Be^{-\alpha_n x}$ ?

Thank you for your effort and your time!

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    @E.Joseph Thanks a lot for the nice editing ! – 2017-01-02
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    *Full answer:* use the identities $$e^{\alpha_nx}=\cosh(\alpha_nx)+\sinh(\alpha_nx)\qquad e^{-\alpha_nx}=\cosh(\alpha_nx)-\sinh(\alpha_nx)$$ – 2017-01-02
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    @Did your approach works! In this case A' = (A+B) and B'=(A-B), right ? – 2017-01-02
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    Exactly. No need to solve a linear system (at the risk of solving it wrongly, as happens below). – 2017-01-02
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    @Did yes, your're right! Thanks a lot !! – 2017-01-02
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    How do you define the hyperbolic functions ? – 2017-01-02

3 Answers 3

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We don't have to be explicit here: $\cosh \alpha_n x$ and $\sinh\alpha_n x$ are independent linear combinations of the linearly independent vectors $\mathrm e^{\alpha_n x}$ and $\mathrm e^{-\alpha_n x}$, hence they span the same subspace.

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Note that $$ A'\cosh(ax)+B'\sinh(ax)=\frac{A'+B'}{2}e^{ax}+\frac{A'-B'}{2}e^{-ax}. $$ So $\frac{A'+B'}{2}=A$ and $\frac{A'-B'}{2}=B$, giving $A'=A+B$ and $B'=A-B$.

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    see that your answer is wrong. just put your answer in the equilities you wrote. – 2017-01-02
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    @kolobokish why??? – 2017-01-02
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    See that if $A'$ and $B'$ are as above, that you'll not get $A$. You'll get $\frac{A}{2}$. – 2017-01-02
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    @kolobokish Thanks. Was a quick fix. – 2017-01-02
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    Now that's ok.) Happy New Year. – 2017-01-02
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It seem to it should be like this. $$Ae^{\alpha x}+Be^{-\alpha x} = Ae^{\alpha x}+ Ae^{-\alpha x}+ (B-A)e^{-\alpha x} = 2Ash(\alpha x)+ (B-A)e^{-\alpha x}$$ But we have that $e^{-x} = ch(x)-sh(x)$. So we will have. $$Ae^{\alpha x}+Be^{-\alpha x} = 2Ash(\alpha x) + (B-A)(ch(\alpha x)-sh(\alpha x)) = (A+B)sh(\alpha x)+ (B-A)ch(\alpha x)$$.

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    Thanks for your answer ! You are using a similar concept than the comment form @Did . – 2017-01-02