# on 12-Nov-2023 (Sun)

#### Flashcard 7597061180684

Question

$$e^θ = \cosh θ + \sinh θ$$

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Euler's Equation in Complex number set

#### Flashcard 7597070617868

Question

Depending on the solution of the auxiliary equation, the solution of the corresponding 2nd Linear & homogeneous ODE is different: $$\displaystyle m=\frac{-b\pm\sqrt{\Delta}}{2a}, \Delta\equiv b^{2}-4ac$$

[Answer the normal solutions and their alternative form]

1. $$\Delta>0\Rightarrow$$ 2 real roots: $$m_{1},m_{2}$$, both $$y_{1}=e^{m_{1}x},y_{2}=e^{m_{2}x}$$ works. So the general solution is $$y_{general}=Ae^{m_{1}x}+Be^{m_{2}x}$$
- Alternative form: Because of the Euler's Equation in $$\mathbb{C}$$, $$y(x)=e^{px}[\tilde{A}\cosh(qx)+\tilde{B}\sinh(qx)]$$, where $$p,q$$ are $$\displaystyle m=\underbrace{- \frac{b}{2a}}_p\pm \underbrace{\frac{\sqrt{\Delta}}{2a}}_{q}$$.
- Why we need this form? When the initial condition includes $$y(0)$$, the alt form would be much easier
2. $$\Delta<0\Rightarrow$$ 2 complex roots: $$m_{1},m_{2}\in\mathbb{C}$$, so the general solution is $$y(x)=Ce^{m_{1}x}+De^{m_{2}x}$$
- Alternative form: Because of the Euler's Equation, $$y(x)=e^{m_{r}x}[\tilde{C}\cos(m_{i}x)+\tilde{D}\sin(m_{i}x)]$$, where $$m_{1,2}=m_{r}\pm im_{i}$$.
3. $$\Delta=0\Rightarrow$$ one $$m$$ only! so the general solution is $$y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}$$
Depending on the solution of the auxiliary equation, the solution of the ODE is different: $$m=\frac{-b\pm\Delta}{2a}, \Delta\equiv b^{2}-4ac$$ 1. $$\Delta>0\Rightarrow$$ 2 real roots: $$m_{1},m_{2}$$, both $$y_{1}=e^{m_{1}x},y_{2}=e^{m_{2}x}$$ works. So the general solution is $$y_{general}=Ae^{m_{1}x}+Be^{m_{2}x}$$ - Alternative form: Because of the Euler's Equation in $$\mathbb{C}$$, $$y(x)=e^{px}[\tilde{A}\cosh(qx)+\tilde{B}\sinh(qx)]$$, where $$p,q$$ are $$\displaystyle m=\underbrace{- \frac{b}{2a}}_p\pm \underbrace{\frac{\sqrt{\Delta}}{2a}}_{q}$$. - Why we need this form? When the initial condition includes $$y(0)$$, the alt form would be much easier 2. $$\Delta<0\Rightarrow$$ 2 complex roots: $$m_{1},m_{2}\in\mathbb{C}$$, so the general solution is $$y(x)=Ce^{m_{1}x}+De^{m_{2}x}$$ - Alternative form: Because of the Euler's Equation, $$y(x)=e^{m_{r}x}[\tilde{C}\cosh(m_{i}x)+\tilde{D}\sinh(m_{i}x)]$$, where $$m_{1,2}=m_{r}\pm im_{i}$$. 3. $$\Delta=0\Rightarrow$$ one $$m$$ only! so the general solution is $$y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}$$