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If an ODE is linear and homogeneous, then the "Principle of Superposition" appears: If \(y_{1}\) and \(y_{2}\) both solve an ODE, then, \(y_{3}=\alpha y_{1}+\beta y_{2}\) is also a solution.

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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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Cauchy–Euler equation is an non-constant coefficients 2nd order linear homogeneous ODE whose form likes \(x^{2}y''+bxy'+cy=0\)

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Cauchy–Euler equation is an non-constant coefficients 2nd order linear homogeneous ODE whose form likes \(x^{2}y''+bxy'+cy=0\)

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If the ODE is like or can be transformed to \(x^{2}y''+bxy'+cy=0\) then it can use the "Euler Equation" method: 1. Try ansatz: \(y=x^{p}\), \(p\) is an unknown constant. 2. Plug it in ODE to get the auxiliary equation \(p(p-1)+ap+b=0\). 3. Using the solution of the auxiliary equation \(p_{1},p_{2}\) to solve the ODE: \(y_{general}(x)=Ax^{p_{1}}+Bx^{p_{2}}\) , where \(A,B=\) constant. - If \(p_{1}=p_{2}=p\), claim: \(y_{general}(x)=(A+B\ln x)x^{p}\)

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For the non-constant coefficients linear homogeneous ODE, there are only two solution methods: - If the ODE is like \(x^{2}y''+bxy'+cy=0\), then it can be fully solved: "Euler Equation" Method - Otherwise, need "Red of Order"