Edited, memorised or added to reading queue

on 09-Nov-2023 (Thu)

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

Flashcard 7597038374156

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

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

Principle of Superposition for ODE's solutions
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.






Annotation 7597063802124

The solution method for Constant-coefficients 2nd order linear homogeneous ODE

The general form of 2nd-order linear homogeneous ODE is \(a_{2}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y(x)=0\)
If \(a_{2}(x),a_{1}(x),a_{0}(x)\) are constant functions:
1. Try ansatz: \(y=e^{mx}\).
2. To find \(m\), plug \(y = e^{mx}\) into ODE. Rearrange the equation and get the auxiliary equation: \(\underbrace{(am^{2}+bm+c)}_{\text{Auxiliary Equation}}e^{mx}=0\)
3. Using the solution of the auxiliary equation to solve the ODE.
statusnot read reprioritisations
last reprioritisation on suggested re-reading day
started reading on finished reading on



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






Annotation 7597079268620

Definition of Cauchy–Euler equation

Cauchy–Euler equation is an non-constant coefficients 2nd order linear homogeneous ODE whose form likes \(x^{2}y''+bxy'+cy=0\)
statusnot read reprioritisations
last reprioritisation on suggested re-reading day
started reading on finished reading on



Flashcard 7597081103628

Question

Cauchy–Euler equation is [...]

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

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

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






Flashcard 7597082676492

Question

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

Answer
Cauchy–Euler equation

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

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






Annotation 7597085822220

The "Euler Equation" Method for non-constant-coefficients 2nd order linear homogeneous ODE

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: \(y_{general}(x)=(A+B\ln x)x^{p}\)
statusnot read reprioritisations
last reprioritisation on suggested re-reading day
started reading on finished reading on



Flashcard 7597087657228

Question
If the ODE is like or can be transformed to \(x^{2}y''+bxy'+cy=0\) then it can use the "Euler Equation" method: [...]
Answer

1. Try ansatz: \(y=x^{p}\), \(p\) is an unknown constant.
2. Plug it in ODE to get the auxiliary equation \(p(p-1)+bp+c=0\).
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: \(y_{general}(x)=(A+B\ln x)x^{p}\)

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

The solution method for non-constant-coefficients 2nd order linear homogeneous ODE by &quot;Euler Equation&quot; Method
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: \(y_{general}(x)=(A+B\ln x)x^{p}\)






Annotation 7597095259404

The solution method for non-constant-coefficients 2nd order linear homogeneous ODE
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"
statusnot read reprioritisations
last reprioritisation on suggested re-reading day
started reading on finished reading on



Flashcard 7597097618700

Question
For the non-constant coefficients linear homogeneous ODE, there are only two solution methods: [Answer the method name and their applied condition]
Answer

- 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"

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

The solution method for non-constant-coefficients 2nd order linear homogeneous ODE
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"