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How can you prove this formula:

$C_1 e^{(\alpha + i\beta) t} + C_2 e^{(\alpha - i\beta)t}=Ke^{\alpha t}\cos {(\beta t + \phi)}$

This gives $x(t)$ in the second-order differential equation for an underdamped system with the characteristic equation $\lambda^2 + A\lambda + B = 0$ which gives $\lambda_{1,2} = \alpha \pm i\beta = \frac {A \pm \sqrt{A^2 - 4B}} 2$, and $C_1$, $C_2$ and $K$ are constants.

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You want to prove that that DE has the characteristic equation $\lambda^2+A \lambda +B$? –  Alyosha May 21 '13 at 15:51

1 Answer 1

up vote 2 down vote accepted

$$e^{ix}=\cos(x)+i\sin(x)$$ Leading to: $$C_1e^{\alpha t} e^{i \beta t}+C_2e^{\alpha t} e^{-i \beta t}=e^{\alpha t} (C_1 \cos(\beta t)+iC_1 \sin(\beta t)+C_2 \cos(\beta t)-iC_2 \sin(\beta t))$$

$$=e^{\alpha t} ((C_1+C_2) \cos(\beta t)-(iC_2-iC_1) \sin(\beta t))$$

$$=e^{\alpha t} (A \cos(\beta t)-B \sin(\beta t))$$

Now, any $(A \cos(\beta t)-B \sin(\beta t))$ can be written in the form $K \cos (\beta t+ \phi)$:

$$K \cos (\beta t+ \phi)=(K\cos(\phi)) \cos(\beta t)-(K\sin(\phi)) \sin(\beta t)$$ As cosine and sine are orthogonal, you can equate the coefficients:

$$K\cos(\phi)=A, K\sin(\phi)=B \Rightarrow \phi=\arctan(\frac{B}{A})$$ Has a unique solution $\frac{-\pi}{2}<\phi<\frac{\pi}{2}$, and $$K^2(\cos(\phi)^2+\sin(\phi)^2)=A^2+B^2 \Rightarrow K=\sqrt{A^2+B^2}$$

So

$$C_1e^{\alpha t} e^{i \beta t}+C_2e^{\alpha t} e^{-i \beta t}=e^{\alpha t}\sqrt{A^2+B^2}\cos(\beta t+ \arctan(\frac{B}{A}))$$

With

$$A=C_1+C_2$$ $$B=iC_2-iC_1$$

Finally,

$$C_1e^{\alpha t} e^{i \beta t}+C_2e^{\alpha t} e^{-i \beta t}=2e^{\alpha t} \sqrt{C_1C_2} \cos(\beta t+i \tanh^{-1}(\frac{C_2-C_1}{C_2+C_1}))$$

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Thanks, Alyosha. I would really like to vote your answer up but I don't have enough reputation. But I will as soon as I get some. –  Mohammad Sanei May 21 '13 at 17:36
    
That's fine, it's just that you deleted the bit in the question when you asked about this explicitly, so I thought I'd answered an unimportant part of your question. –  Alyosha May 21 '13 at 18:08

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