Contents
- 1. Linear first-order ODE
- 2. Exact equation
- 3. Constant coefficient Second order linear ODE
- 4. Variable coefficient Second order linear ODE
- 5. Variation of parameter과 second order linear ODE의 적용
- 6. Harmonic Oscillation, Resonance
- 7. First and second order linear ODE
- 8. Existence and Uniqueness Theorem
- 9. Laplace transform 1
- 10. Laplace transform 2
- 11. Laplace transform 3
- 12. Laplace transform 4
1. Linear first-order ODE ¶
from 최정환 https://www.youtube.com/watch?v=J2FJVHchcu4&list=PLL3t9Nt4HrfvpCtvSnkH6Uw0rr169Pzkv&index=23
(a)
미분방정식
$\displaystyle y'=f(x)$
의 해는,
$\displaystyle y=\int f(x)dx$
이것은 고등학교 때 배움.
미분방정식
$\displaystyle y'=f(x)$
의 해는,
$\displaystyle y=\int f(x)dx$
이것은 고등학교 때 배움.
(b)
$\displaystyle y'=y$
$\displaystyle y=ce^x$
easy
$\displaystyle y'=y$
$\displaystyle y=ce^x$
easy
(c)
$\displaystyle y'=ay$
$\displaystyle y=ce^{ax}$
easy
$\displaystyle y'=ay$
$\displaystyle y=ce^{ax}$
easy
(d)
$\displaystyle y'=ay+b$
sol. (i)
$\displaystyle y'=a(y+\frac{b}a)$
$\displaystyle Y=y+\frac{b}a$
$\displaystyle Y'=aY$
sol. (ii)
$\displaystyle y'+(-a)y=b$
$\displaystyle e^{-ax}y'+(-a)e^{-ax}y=be^{-ax}$
좌변은 $\displaystyle =(e^{-ax}y)'$
$\displaystyle e^{-ax}y=-\frac{b}{a}e^{-ax}+c$
$\displaystyle y=-\frac{b}a+ce^{ax}$
$\displaystyle y'=ay+b$
sol. (i)
$\displaystyle y'=a(y+\frac{b}a)$
$\displaystyle Y=y+\frac{b}a$
$\displaystyle Y'=aY$
sol. (ii)
$\displaystyle y'+(-a)y=b$
$\displaystyle e^{-ax}y'+(-a)e^{-ax}y=be^{-ax}$
좌변은 $\displaystyle =(e^{-ax}y)'$
$\displaystyle e^{-ax}y=-\frac{b}{a}e^{-ax}+c$
$\displaystyle y=-\frac{b}a+ce^{ax}$
(e)
$\displaystyle y'=ay+b(x)$
(d)의 (i)방식으로는 너무 복잡해진다.
(d)의 (ii)방식으로 하면,
$\displaystyle e^{-ax}y'-ae^{-ax}y=b(x)e^{-ax}$
좌변은 $\displaystyle =(e^{-ax}y)'$
$\displaystyle y=e^{ax}\int b(x)e^{-ax}dx$
$\displaystyle y'=ay+b(x)$
(d)의 (i)방식으로는 너무 복잡해진다.
(d)의 (ii)방식으로 하면,
$\displaystyle e^{-ax}y'-ae^{-ax}y=b(x)e^{-ax}$
좌변은 $\displaystyle =(e^{-ax}y)'$
$\displaystyle y=e^{ax}\int b(x)e^{-ax}dx$
(f) 가장 일반적인 형태
$\displaystyle y'=a(x)y+b(x)$ or
$\displaystyle y'+a(x)y=b(x)$
$\displaystyle e^{\int a(x)dx}$ 를 곱하면
$\displaystyle e^{\int a(x)dx}y'+a(x)e^{\int a(x)dx}y=b(x)e^{\int a(x)dx}$
좌변이 $\displaystyle \left[e^{\int a(x)dx}y\right]'$
$\displaystyle e^{\int a(x)dx}y=\int\left[b(x)e^{\int a(x)dx}\right]dx$
$\displaystyle y=e^{-\int a(x)dx}\int\left[b(x)e^{\int a(x)dx}\right]dx$
$\displaystyle y'=a(x)y+b(x)$ or
$\displaystyle y'+a(x)y=b(x)$
$\displaystyle e^{\int a(x)dx}$ 를 곱하면
$\displaystyle e^{\int a(x)dx}y'+a(x)e^{\int a(x)dx}y=b(x)e^{\int a(x)dx}$
좌변이 $\displaystyle \left[e^{\int a(x)dx}y\right]'$
$\displaystyle e^{\int a(x)dx}y=\int\left[b(x)e^{\int a(x)dx}\right]dx$
$\displaystyle y=e^{-\int a(x)dx}\int\left[b(x)e^{\int a(x)dx}\right]dx$
2. Exact equation ¶
$\displaystyle y'=f(x,y)=\frac{-M(x,y)}{N(x,y)}$
$\displaystyle \Leftrightarrow M(x,y)+N(x,y)y'=0$
$\displaystyle \Leftrightarrow M(x,y)+N(x,y)\frac{dy}{dx}=0$
$\displaystyle \Leftrightarrow M(x,y)dx+N(x,y)dy=0$
$\displaystyle \Leftrightarrow M(x,y)+N(x,y)y'=0$
$\displaystyle \Leftrightarrow M(x,y)+N(x,y)\frac{dy}{dx}=0$
$\displaystyle \Leftrightarrow M(x,y)dx+N(x,y)dy=0$
is exact if $\displaystyle \exists F(x,y)$ such that $\displaystyle F_x=M$ and $\displaystyle F_y=N$
Thm.
If $\displaystyle M,N,M_y,N_x$ are all continuous, then
$\displaystyle M_y=F_{xy}=F_{yx}=N_x$
suppose $\displaystyle M_y=N_x$
If $\displaystyle M,N,M_y,N_x$ are all continuous, then
$\displaystyle Mdx+Ndy=0$
is exact iff$\displaystyle M_y=N_x$ .
Pf. $\displaystyle \exists F$ s.t. $\displaystyle F_x=M,F_y=N$$\displaystyle M_y=F_{xy}=F_{yx}=N_x$
suppose $\displaystyle M_y=N_x$
3. Constant coefficient Second order linear ODE ¶
2nd order linear homogeneous equation with constant coefficients
$\displaystyle Ay''+By'+Cy=0$ (A,B,C 상수)
$\displaystyle y=e^{\lambda x},$ λ 상수
$\displaystyle y'=\lambda e^{\lambda x}$
$\displaystyle y''=\lambda^2 e^{\lambda x}$
그러면 본 식은
$\displaystyle A\lambda^2 e^{\lambda x}+B\lambda e^{\lambda x}+Ce^{\lambda x}=0$
$\displaystyle (A\lambda^2+B\lambda+C)e^{\lambda x}=0$
왼쪽의 $\displaystyle (A\lambda^2+B\lambda+C)$ 을 characteristic equation이라 함.
case 1. $\displaystyle B^2-4AC>0$
이차방정식이 두 실근 $\displaystyle \lambda_1\ne\lambda_2$ 를 가진다.
$\displaystyle y'=\lambda e^{\lambda x}$
$\displaystyle y''=\lambda^2 e^{\lambda x}$
그러면 본 식은
$\displaystyle A\lambda^2 e^{\lambda x}+B\lambda e^{\lambda x}+Ce^{\lambda x}=0$
$\displaystyle (A\lambda^2+B\lambda+C)e^{\lambda x}=0$
왼쪽의 $\displaystyle (A\lambda^2+B\lambda+C)$ 을 characteristic equation이라 함.
case 1. $\displaystyle B^2-4AC>0$
이차방정식이 두 실근 $\displaystyle \lambda_1\ne\lambda_2$ 를 가진다.
생략..
4. Variable coefficient Second order linear ODE ¶
$\displaystyle y''+p(x)y'+q(x)y=0$
Suppose a solution $\displaystyle y_1$ is given.
Let $\displaystyle y_2=y_1v(x),\;v(x)\ne C$
Put $\displaystyle y_2=y_1v(x)$ into the given equation.
$\displaystyle (y_1v)''+P(y_1v)'+qy_1v=0$
$\displaystyle (y_1v)'=y_1'v+y_1v'$
$\displaystyle (y_1v){' '}=y_1{' '}v+2y_1'v'+y_1v{' '}$
Suppose a solution $\displaystyle y_1$ is given.
Let $\displaystyle y_2=y_1v(x),\;v(x)\ne C$
Put $\displaystyle y_2=y_1v(x)$ into the given equation.
$\displaystyle (y_1v)''+P(y_1v)'+qy_1v=0$
$\displaystyle (y_1v)'=y_1'v+y_1v'$
$\displaystyle (y_1v){' '}=y_1{' '}v+2y_1'v'+y_1v{' '}$
이하skip
7. First and second order linear ODE ¶
First order ODE의 해법
2계 선형 상미분 방정식의 수리적 해석
https://www.youtube.com/watch?v=UYvnJDhAUt0&index=31&list=PLL3t9Nt4HrfvpCtvSnkH6Uw0rr169Pzkv
2계 선형 상미분 방정식의 수리적 해석
https://www.youtube.com/watch?v=UYvnJDhAUt0&index=31&list=PLL3t9Nt4HrfvpCtvSnkH6Uw0rr169Pzkv