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

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
?

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

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