解聯立微分方程式 摺積定理

1.聯立微分方程式(x1)' = 4(x1) - 2(x2) + e^t(x2)' = -2(x1) + (x2)Ans: (1) 簡化運算: x=x1, y=x2{x}'.[.4 -2].{x}.{e^t}
{y} =[-2 .1]*{y}+{0..}令X={x}
....{y}A=[.4 -2]
..[-2 .1]B={e^t}
..{0..}=> X'-A*X=B|A|=4-4=0 => 不能使用矩陣法
(2) 改用微分法: 再微分一次x'=4x-2y+e^t.....(a)y'=-2x+y.....(b)(a)' & (b)':x"=4x'-2y'+e^ty"=-2x'+y'=-2(4x-2y+e^t)+(-2x+y)=-8x+4y-2e^t-2x+y=-10x+5y-2e^t=5(-2x+y)-2e^t=5*y'-2e^t=> y"-5y'=-2e^t特徵方程式:0=m^2-5m=m(m-5)m=0,5齊次解為: yh=a+b*e^5ta,b=常數
(3) 求特殊解yp=yp'=yp"=c*e^typ"+yp'=2c*e^t=-2*e^t => c=-1=> yp=-e^t
(4) 求通解y(t)=yh+yp=a+b*e^5t-e^t.....(c)=ans
(5) 求另外一變數(c),(a):x'=4x-2y+e^t=4x-2(a+b*e^5t-e^t)+e^t=4x-2a-2b*e^5t+3e^t=> x'-4x=-2a-2b*e^5t+3e^t特徵方程式:0=m-4 => m=4齊次解為: xh=f*e^4t(6) 求特殊解xp=g*e^5t+h*e^t+kxp'=5g*e^5t+h*e^txp'-4xp=5g*e^5t+h*e^t-4(g*e^5t+h*e^t+k)=g*e^5t-3h*e^t-4k=-2a-2b*e^5t+3e^t=> g=-2b, h=-1, k=1/2=> xp=-2b*e^5t-h*e^t+1/2
(7) 求通解x(t)=xh+xp=f*e^4t-2b*e^5t-h*e^t+1/2.....ans

2.摺積定理: 請問sint*sint 如何用摺積公式求解出...就是積分公式那個，看得懂，但是不會用呀....Ans: 褶積分定理(Convolution Theory) => A=L[f(t)*g(t)]=L{∫<0~t>g(R)*f(t-R)dR}=L{∫sinR*sin(t-R)dR}=L{∫sinR*(sinR*cost-sint*cosR)dR}=L{∫sin^2R*cost*dR-∫sint*cosR*sinR*dR}=L{∫cost*(1-cos2R)dR/2-∫sint*sinR*d(sinR)}=L{∫cost*dR/2-∫cost*cos2R*dR/2-sint*sin^2R/2}=L{R*cost/2-cost*cos2R/4-sint*sin^2R/2}.....R=0~t=L{(t/2)*cost-cost(cos2t-1)/4-(sint*sin2t)/2}=L{0.5*t*cost-0.25*cost*cos2t+0.25cost-0.5*sint*sin2t}=L{0.5*t*cost-0.25*cost*cos2t+0.25cost-sint*sint*cost}=L{cost*[0.5t-0.25cos2t+0.25-sint*sint]} =G(s)*H(s)=L(cost)*L[0.5t-0.25cos2t+0.25-sint*sint]=> G(s)=L(cost)=1/(s^2+1)=> H(s)=L[0.5t-0.25cos2t+0.25-sint*sint]=L(0.5t-0.25cos2t+0.25)-L(sint*sint)=1/2s^2-2/4(s^2+4)+1/4s-A=> A=G(s)*H(s)=[1/(s^2+1)]*[1/2s^2-1/2(s^2+4)+1/4s-A]=[1/(s^2+1)]*[1/2s^2-1/2(s^2+4)+1/4s]-A/(s^2+1)=> A[1+1/(s^2+1)]=[1/(s^2+1)]*[1/2s^2-1/2(s^2+4)+1/4s]A=[1/(s^2+1)]*[1/2s^2-1/2(s^2+4)+1/4s]*(s^2+1)/(s^2+2)=[1/2s^2-1/2(s^2+4)+1/4s]/(s^2+2).....ans

到下面的網址看看吧

▶▶http://misshare168.pixnet.net/blog/post/86950298

到下面的網址看看吧

▶▶http://misshare168.pixnet.net/blog/post/86950298

1. (x1)' = 4(x1) - 2(x2) + e^t---(1)
(x2)' = -2(x1) + (x2)-----------(2)
(1)+2(2)----> (x1+2x2)'=e^t---(3) ==> x1+2[x2] =e^t+c1----(4)
From (4) 2[x2]=-x1+e^t+c1 ---(5), plug into (1) -->
(x1)'=4{x1]+[x1]-e^t-c1+e^t ---(6), a linear equation for x1(t) ==>
x1(t)=(c1)/5+(c2) e^(5t), ---(7),where c1, and c2 are arb. constants.-->plug (7)into
(4), x2(t)=(1/2){-x1(t)+e^t+c1} ---(8). 即得一般解[(7),(8)].
沒有始初條件下，Laplace transform 法徒增麻煩。

2. By definition, sin(t)摺積 sin(t)=int[s=0 to t] {sin(t-s)sin(s) ds}
=int[s=0 to t] {[sin(t)cos(s)-cos(t)sin(s)]sin(s) ds}
=sin(t) int[s=0 to t] {cos(s)sin(s) ds} -cos(t) int[s=0 to t] {sin(s)sin(s) ds}
=sin(t) int[s=0 to t] {(1/2)sin(2s) ds} -cos(t) int[s=0 to t] {(1/2)[1-cos(2s)]ds}
=(1/4)sin(t)[1-cos(2t)]-cos(t)[(1/2)t-(1/4)sin(2t)].....[留給大大化簡]