Edited, memorised or added to reading queue

on 12-Nov-2023 (Sun)

Do you want BuboFlash to help you learning these things? Click here to log in or create user.

Flashcard 7597061180684

Question
由于该公式在\(x\)为复数时仍然成立,所以也有人将这一更通用的版本称为欧拉公式:[...]
Answer
\(e^θ = \cosh θ + \sinh θ\)

statusnot learnedmeasured difficulty37% [default]last interval [days]               
repetition number in this series0memorised on               scheduled repetition               
scheduled repetition interval               last repetition or drill

Euler's Equation in Complex number set
由于该公式在\(x\)为复数时仍然成立,所以也有人将这一更通用的版本称为欧拉公式:\(e^{iθ} = \cosθ + i\sin θ \iff e^θ = \cosh θ + \sinh θ\)







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]

Answer
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 \(y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}\)

statusnot learnedmeasured difficulty37% [default]last interval [days]               
repetition number in this series0memorised on               scheduled repetition               
scheduled repetition interval               last repetition or drill

The relations between the solution of an ODE and the solution of the ODE's auxiliary equation
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 \(y(x)=Ae^{mx}+Bxe^{mx}=(A+Bx)e^{mx}\)