D notation/operator

$D[(D+2t)[t^3-8]]$

$(D+2t)[D[t^3-8]]$

$D(D+2t) \ne (D+2t)D$
내부에 t가 있으면 교환법칙 성립안함.
$D(D+2) = (D+2)D=D^2+2D$
내부에 t가 없으면 교환가능

$(D+2)[(D-1)[t^3-8]]$

$(D-1)[(D+2)[t^3-8])$

$(D^2+D-2)[t^3-8]$

$(D+2)(D-1)=(D-1)(D+2)$
$(D+2)(D-1)=(D^2+D-2)$

System D.E.

$x'(t)=3x(t)-4y(t)+1 \brace y'(t)=4x(t)-7y(t)+10t$

$\begin{cases} x'(t) &= 3x(t)-4y(t)+1 \\ y'(t) &= 4x(t)-7y(t)+10t \\ \end{cases} \\ y''(t) = 4x'-7y'+10 \\ 4x'=12x-16y+4\\ 3y'=12x-21y+30t\\ 4x'-3y'=5y+4-30t\\ 4x' = 3y'+5y+4-30t \\ y'' =3y'+5y+4-30t-7y'+10 \\ y'' +4y' -5y = -30t + 14 \\ \text{found 2nd order LDE.}$
$y_h: r^2+4r-5=0, (r-1)(r+5)=0\\ r=1 \lor -5 \\ y_h: e^t, e^{-5t} \\ y_p form: At+B\\ y'=A, y''=0\\ y'' +4y' -5y = -30t + 14 \\ 0+4A-5At-5B=-30t+14\\ A=6, \quad 24-5B=14, \quad B=2\\ y_p = 6t+2\\ y = De^t+Ee^{-5t}+6t+2\\$

$\text{연립방정식에서 x DE를 y처럼 구할수도 있지만 \\ y를 대입해서 구한다.}\\ y'(t) = 4x(t)-7y(t)+10t\\ 4x(t) = y'+7y-10t\\ =De^t-5Ee^{-5t}+6+7De^t+7Ee^{-5t}+42t+14-10t\\ =2Ee^{-5t}+8De^t+32t+20\\ x(t) = \frac 1 2Ee^{-5t}+2De^t+8t+5$

Alt. Use D operator (D 연산자를 이용한 방식)
$\begin{cases} x'(t) &= 3x(t)-4y(t)+1 \\ y'(t) &= 4x(t)-7y(t)+10t \\ \end{cases} \\ \begin{cases} x' -3x +4y &= 1 \\ -4x +y'+7y&= 10t \\ \end{cases} \\ \begin{cases} (D-3)(x) +4y &= 1 \\ -4x +(D+7)(y)&= 10t \\ \end{cases} \\ \begin{cases} 4(D-3)(x) +16y &= 4 \\ (D-3)(-4x) +(D-3)(D+7)(y)&= (D-3)(10t) \\ \end{cases} \\$
$16y+(D-3)(D+7)(y) = (D-3)(10t)+4\\ 16y+(D^2+4D-21)y = 10-30t+4\\ 16y+y''+4y'-21y=-30t+14\\ y''+4y'-5y=-30t+14 \\$
위와 동일한 2계 미방을 구했다. 이하 생략.

$y''+4y'-5y=14-30t$

$(D^2+4D-5) y = 14-30t\\ D^2(D^2+4D-5) y = D^2(14-30t)=0\\ D^2(D^2+4D-5)=0\\ D^2(D+5)(D-1)=0\\ D=0(repeated), 1, -5\\ y form: C, Ct, e^t, e^{-5t}\\ y_p form : At+B (\text{C, Ct same form.})\\ input : y_p \\ y''+4y'-5y=14-30t \\ 0+4A-5At-5B=14-30t\\ A=6, B=2\\ y_p = 6t+2\\ y = C_1e^t+C_2e^{-5t}+6t+2$

system of differential equation.

Q $x'+y'+2x=0 \brace x'+y'-x-y=sin(t)$

$\begin{cases} x'+y'+2x=0 \quad &(1)\\ x'+y'-x-y=sin(t) \quad &(2) \end{cases} \\ \begin{cases} (D+2)x+Dy=0 \\ (D-1)x+(D-1)y=sin(t) \end{cases}\\ substract \begin{cases} (D-1)(D+2)x+(D-1)Dy=(D-1)0 \\ (D+2)(D-1)x+(D+2)(D-1)y=(D+2)sin(t) \end{cases}\\ (D^2-D)y-(D^2+D-2)y=-(D+2)sin(t)\\ y''-y'-(y''+y'-2y)=-(cos(t)+2sin(t))\\ -2y'+2y=-cos(t)-2sin(t)\\$

$2y'-2y=cos(t)+2sin(t)\\ (2D-2)y=cos(t)+2sin(t)\\ y_h : 2r-2=0, r=1 \\ y_h = C_1e^{t}\\ y_p form : y_p = Acos(t)+Bsin(t)\\ y_p' = -Asin(t)+Bcos(t) \\ 2y'-2y=cos(t)+2sin(t) \\ -2Asin(t)+2Bcos(t)-2Acos(t)-2Bsin(t) = cos(t)+2sin(t) \\ (2B-2A)cos(t)+(-2A-2B)sin(t) = cos(t)+2sin(t)\\ 2B-2A=1, -2A-2B=2 \\ -4A=3, A=-\frac 3 4 , 2B+\frac 3 2 = 1, 2B=-\frac 1 2\\ B=-\frac 1 4\\ \therefore y = C_1e^{t}-\frac 3 4 cos(t)-\frac 1 4 sin(t) \\$
Alternative.
$-2y'+2y=-cos(t)-2sin(t)\\ y'-y=\frac 1 2 cos(t)+sin(t)\\ Int.Factor=e^{\int -1 dt}=e^{-t}\\ (e^{-t}y)' = e^{-t}(\frac 1 2 cos(t)+sin(t))\\ e^{-t}y = \int e^{-t}(\frac 1 2 cos(t)+sin(t)) dt\\ =\frac 1 2 \int e^{-t}cos(t)dt+\int e^{-t} sin(t)dt\\ \int e^{-t}cos(t) dt = e^{-t}sin(t)-(-e^{-t})(-cos(t))+\int e^{-t}(-cos(t))dt\\ \int e^{-t}cos(t) dt = \frac 1 2 e^{-t}(sin(t)-cos(t)) \\ \int e^{-t}sin(t) dt = e^{-t}(-cos(t))-(-e^{-t})(-sin(t))+\int e^{-t}(-sin(t))dt\\ \int e^{-t}sin(t) dt = -\frac 1 2 e^{-t}(cos(t)+sin(t))\\ e^{-t}y = \frac 1 4 e^{-t}(sin(t)-cos(t))-\frac 1 2 e^{-t}(cos(t)+sin(t))+C_1\\ y = \frac 1 4 (sin(t)-cos(t))-\frac 1 2(cos(t)+sin(t))+C_1e^t\\ \therefore y = -\frac 3 4 cos(t)-\frac 1 4 sin(t)+C_1e^t \quad \text{same result. good.}\\ y' = C_1e^{t}+\frac 3 4 sin(t)-\frac 1 4 cos(t) \\$

And…

• Try to use EQ. (1)
  $\begin{aligned} &x'+y'+2x=0 \\ &x'+C_1e^{t}+\frac 3 4 sin(t)-\frac 1 4 cos(t)+2x=0\\ &x'+2x=-C_1e^{t}-\frac 3 4 sin(t)+\frac 1 4 cos(t)\\ &(D+2)x=-C_1e^{t}-\frac 3 4 sin(t)+\frac 1 4 cos(t)\\ &x_h : r+2=0, r=-2\\ &x_h = E e^{-2t}\\ &x_p form = Ae^{t}+Bcos(t)+Csin(t) \\ &x_p' = Ae^{t}-Bsin(t)+Ccos(t) \\ &x'+2x=-C_1e^{t}-\frac 3 4 sin(t)+\frac 1 4 cos(t)\\ &Ae^{t}-Bsin(t)+Ccos(t)+2Ae^{t}+2Bcos(t)+2Csin(t)\\ &=-C_1e^{t}-\frac 3 4 sin(t)+\frac 1 4 cos(t)\\ &3Ae^{t}-(B-2C)sin(t)+(2B+C)cos(t)=-C_1e^{t}-\frac 3 4 sin(t)+\frac 1 4 cos(t)\\ &3A=-C_1, B-2C=\frac 3 4, 2B+C=\frac 1 4\\ &4B+2C=\frac 2 4, 5B=\frac 5 4, B=\frac 1 4, C=\frac 1 4-\frac 2 4=-\frac 1 4\\ &x_p = -\frac 1 3 C_1e^{t}+\frac{1}{4}cos(t)-\frac 1 {4}sin(t)\\ & \therefore x(t) = Ee^{-2t}-\frac 1 3 C_1e^{t}+\frac{1}{4}cos(t)-\frac 1 {4}sin(t)\\ \end{aligned}$
  But… $e^{-2t}$ ??? is Not the solution.

Alternative.
$\begin{aligned} &x'+2x=-C_1e^{t}-\frac 3 4 sin(t)+\frac 1 4 cos(t)\\ &Int.Factor = e^{\int 2 dt}=e^{2t}\\ &(e^{2t}x)' = -C_1e^{3t}-\frac 3 4 e^{2t}sin(t)+\frac 1 4 e^{2t}cos(t)\\ &e^{2t}x = -C_1\int e^3tdt-\frac 3 4\int e^{2t}sin(t)dt+\frac 1 4 \int e^{2t}cos(t)dt\\ &\int e^{2t}sin(t)dt =e^{2t}(-cos(t))-2e^{2t}(-sin(t))+\int 4e^{2t}(-sin(t))dt \\ &5\int e^{2t}sin(t)dt = e^{2t}(-cos(t)+2sin(t))\\ &\int e^{2t}sin(t)dt = \frac 1 5 e^{2t}(-cos(t)+2sin(t)) \\ &\int e^{2t}cos(t)dt =e^{2t}(sin(t))-2e^{2t}(-cos(t))+\int 4e^{2t}(-cos(t))dt \\ &5\int e^{2t}cos(t)dt = e^{2t}(sin(t)+2cos(t))\\ &\int e^{2t}cos(t)dt = \frac 1 5 e^{2t}(sin(t)+2cos(t)) \\ &\int e^{3t}t dt = \frac 1 3 \int e^u du =\frac 1 3 e^{u} = \frac 1 3 e^{3t}\\ & \therefore e^{2t}x = -C_1\int e^3tdt-\frac 3 4\int e^{2t}sin(t)dt+\frac 1 4 \int e^{2t}cos(t)dt\\ &=-\frac {C_1} 3 e^{3t}-\frac 3 {20} e^{2t}(-cos(t)+2sin(t)) +\frac 1 {20} e^{2t}(sin(t)+2cos(t)) + C_2 \\ &x = -\frac {C_1} 3 e^t-\frac 3 {20}(2sin(t)-cos(t))+\frac 1 {20}(sin(t)+2cos(t))+C_2e^{-2t}\\ &x = -\frac {C_1} 3 e^t-\frac 5 {20}sin(t)+\frac 5 {20}cos(t)+C_2e^{-2t}\\ &x = -\frac {C_1} 3 e^t-\frac 1 {4}sin(t)+\frac 1 {4}cos(t)+C_2e^{-2t}\\ &e^{-2t} \text{ found again } ?? &\quad \\ \end{aligned}$

• Try to use EQ. (2)
  $\begin{aligned} &x'+y'-x-y=sin(t)\\ &x'-x+y'-y=sin(t)\\ &x'-x+C_1e^{t}+\frac 3 4 sin(t)-\frac 1 4 cos(t)-(-\frac 3 4 cos(t)-\frac 1 4 sin(t)+C_1e^t) \\ &= sin(t)\\ &x'-x+sin(t)+\frac 1 2 cos(t)=sin(t)\\ &x'-x=-\frac 1 2 cos(t) \\ & IntegrationFactor = e^{-t}\\ &( e^{-t}x)' = -\frac 1 2 e^{-t} cos(t) \\ &e^{-t}x = -\frac 1 2 \int e^{-t} cos(t) dt \\ &e^{-t}x = -\frac 1 2 (\frac 1 2 e^{-t}(sin(t)-cos(t)) +C_3)\\ &x = \frac 1 4 (cos(t) - sint(t))+C_3e^t \end{aligned}$
  alternative.
  $\begin{aligned} &x'-x=-\frac 1 2 cos(t) \\ &2x'-2x=-cos(t)\\ &\text {1. find }x_h\\ & 2x'-2x=0 , \quad (2D-2)x=0\\ & 2r-2=0, r=1 \\ & x_h = C_3e^t \\ & \text{2. find }x_p\\ & x_p = Acost+Bsint\\ & x_p' = -Asint+Bcost \\ &2x'-2x=-cos(t)\\ &2(-Asint+Bcost)-2(Acost+Bsint)=-cos(t)\\ &(2B-2A)cost + (-2A-2B)sin t = -cost \\ &2B-2A=-1, -2A-2B=0, A=-B, 4B=-1, \\ &B=-1/4, A=1/4\\ &\text{3. find } x \\ &x = C_3e^t+\frac 1 4 cos t - \frac 1 4 sin t \end{aligned}$

Q. $x''=y \quad x(0)=3 \quad x'(0)=1 \brace y''=x \quad y(0)=1 \quad y'(0)=-1$

$D^2(x)-y=0 \\ D^2(y)-x=0 \\ D^4(x)-D^2(y)=0\\ D^4(x)-x=0\\ (D^4-1)(x)=0 \\ r^4-1=0 \\ (r^2-1)(r^2+1)=0 \\ r=\pm1 , \quad r=\pm i\\ x(t) = C_1e^{-t}+C_2e^{t}+C_3cos(t)+C_4sin(t)\\ x'(t) = -C_1e^{-t}+C_2e^{t}-C_3sin(t)+C_4cos(t)\\ x''(t) = C_1e^{-t}+C_2e^{t}-C_3cos(t)-C_4sin(t)\\ x(0) = C_1+C_2+C_3=3\\ x'(0)=-C_1+C_2+C_4=1\\ 2C_2+C_3+C_4=4\\ y = D^2(x) = C_1e^{-t}+C_2e^{t}-C_3cos(t)-C_4sin(t)\\ y' = -C_1e^{-t}+C_2e^{t}+C_3sin(t)-C_4cos(t)\\ y(0) = C_1+C_2-C_3=1 \\ y'(0) = -C_1+C_2-C_4=-1 \\ 2C_2-C_3-C_4=0\\ 4C_2=4, C_2=1\\ C_1+C_3=2, C_4-C_1=0, C_1=C_4\\ C_1-C_3=0, C_1=C_3=1\\ \therefore x(t) = e^{-t}+e^{t}+cos(t)+sin(t)\\ y(t) = e^{-t}+e^{t}-cos(t)-sin(t)$

Author: crazyj7@gmail.com