佩蓓 發問時間: 科學數學 · 5 年前

跪求神人大大幫解微分方程(p.s.要詳解喔><) 20點!!

Find Laplace Transform:

# 1. a + bt+ ct^2 (Note that a, b, and c are constants.)

# 2. cos^2 (w t)

# 3. f(t)=minimum{t,1} when 0≦t<∞

# 4. e^t cosh(2t)

Derive inverse transform formula:

# 5. Derive the formula for L^(−1){ 1 / ( s^2 + k^2 )^2 }

by applying the tool L: tf → – F' (s) to the function f(t)=cos(kt).

Hint: Then, combine with the transform of sin(kt).

Find inverse transform:

# 6. 2s^3 ÷ (s^4-1)

# 7. s ÷ [ (s+0.5)^2 +1 ]

# 8. 1 ÷ [ s^3 -s ]

Use Laplace Transform to Solve the ODE:

# 9. y''+ y = 2 cos t ; y(0)=3, y'(0)=4

# 10. y''+ 2y' -3 y = 6e^(-2t) ; y(0)=4, y'(0)=-28

# 11. y''+ y = δ(t -π)-δ(t -2π) ; y(0)=0, y'(0)=1

Hint for #11,12: δ(t -a) → e^(–as), u(t -a) → ( e^(–as) ) / s

Find Laplace Transform:

# 12. f(t) = t u(t-1)

# 13. f(t) = sin t when 2π<t<4π; and f(t)=0 for the remaining t.

2 個解答

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  • 麻辣
    Lv 7
    5 年前
    最佳解答

    公式1: L[t^n]=n!/s^m; m=n+1公式2. L[e^at}=1/(s-a)公式3: L[cos(kt)]=s/(s^2+k^2)公式4. L[sin(kt)]=k/(s^2+k^2)公式5. L[e(at)*f(t)]=F(s-a)公式6. L[t*f(t)]=-F'(s)

    1.Find Laplace Transform:#1.L{a+bt+ct^2}; a,b,c=constants=a/s + b/s^2 + 2c/s^3

    #2.L{cos^2(wt)}=L{[1+cos(2wt)]/2}=L{1}/2+L{cos(2wt)}/2=1/2s + s/2(s^2+4w^2)

    #3.f(t)=minimum{t,1} when 0<=t<∞題意不明

    #4.y4=e^t*cosh(2t)=e^t*AL{A}=L{[e^2t+e^(-2t)]/2}=1/2(s-2)+1/2(s+2)=s/(s^2-4)L{y4}=(s-1)/[(s-1)^2-4]=(s-1)/(s^2-2s-3)=(s-1)/(s+1)(s-3)

    2.Derive inverse transform formula:#5.Derive the formula for £{1/(s^2+k^2)^2}=?A=L{sin(kt)}=k/(s^2+k^2)=k(s^2+k^2)/(s^2+k^2)^2

    L{cos(kt)}=s/(s^2+k^2) B=L{t*cos(kt)}=-[s/(s^2+k^2)]'......Eq.(6)=(2s^2-s^2-k^2)/(s^2+k^2)^2=(s^2-k^2)/(s^2+k^2)^2A-k*B=[k(s^2+k^2)-k(s^2-k^2)]/(s^2+k^2)^2=2k^3/(s^2+k^2)=> 1/(s^2+k^2)=(A-k*B)/2k^3=L{sin(kt)-k*t*cos(kt)}/2k^3=L{sin(t)-t*cos(t)}/2......k=1

    3.Find inverse transform:#6.F6=2s^3÷(s^4-1)=a/(s+1)+b/(s-1)+(cs+d)/(s^2+1)g(s)=2s^3=a(s-1)(s^2+1)+b(s+1)(s^2+1)+(cs+d)(s^2-1)g(1)=2=4b => b=1/2g(-1)=-2=-4a => a=1/2s^3係數: 2=a+b+c => c=1g(0)=0=-a+b-d => d=0£{F6}=£{1/2(s+1) + 1/2(s-1) + s/(s^2+1)}=e^(-t)/2 + e^t/2 + cos(t)

    #7.F7=s÷[(s+0.5)^2+1] L{cos(t)}=s/(s^2+1), L{sin(t)}=1/(s^2+1)A=L{e^(-0.5t)*cos(t)}=(s+0.5)/[(s+0.5)^2+1]=s/[(s+0.5)^2+1] + 0.5/[(s+0.5)^2+1]B=L{e^(-0.5t)*sin(t)}=1/[(s+0.5)^2+1]2A-B=2s/[(s+0.5)^2+1]=> F7=(2A-B)/2=e^(-0.5t)*{cos(t)-sin(t)/2}

    #8.F8=1÷s(s^2-1)=a/s+b/(s+1)+c/(s-1) g(s)=1=a(s^2-1)+bs(s-1)+cs(s+1)g(0)=1=-a => a=-1g(1)=1=2c => c=1/2g(-1)=1=2b => b=1/2£{F8}=£{-1/s + 1/2(s+1) + 1/2(s-1)}=-t + e^(-t)/2 + (e^t)/2

    4.Use Laplace Transform to Solve the ODE:#9.y"+y=2cos(t); y(0)=3, y'(0)=4 L(y"+y)=L(2cos(t))s^2*Y-3s-4+Y=2s/(s^2+1)(s^2+1)Y=2s/(s^2+1)+3s+4Y(s)=2s/(s^2+1)^2 + 3s/(s^2+1) + 4/(s^2+1)=> f(t)=t*sin(t) + 3*cos(t) + 4*sin(t)where L{t*sint}=-{1/(s^2+1)}'=2s/(s^2+1)^2.....Eq.(6)

    2014-10-29 09:01:20 補充:

    #10.y"+2y'-3y=6e^(-2t); y(0)=4, y'(0)=-28

    L(y"+2y'-3y)=L[6e^(-2t)]

    (s^2*Y-4s+28)+2(sY-4)-3Y=6/(s-2)

    (s^2+2s-3)Y=4s-20+6/(s-2)

    Y(s)=4s/(s+3)(s-1) - 20/(s+3)(s-1) + 6/(s+3)(s-1)(s-2)

    =[1/(s-1)+3/(s+3)]+[5/(s-1)-5/(s+3)]+[1/(s-1)-2/(s+2)+1.5/(s+3)]

    =7/(s-1)-0.5/(s+3)-2/(s+2)

    2014-10-29 09:01:39 補充:

    => f(t)=7e^t-0.5e^(-3t)-2e^(-2t)

    2014-10-29 09:22:47 補充:

    #11.y"+y=δ(t-π)-δ(t-2π); y(0)=0, y'(0)=1

    Hint: L{δ(t-a)}=e^(as), L{u(t-a)}=e^(-as)/s

    課本的L{u(t-a)}=e^(as)/s是錯誤的

    L{y"+y}=L{δ(t-π)-δ(t-2π)}

    s^2*Y-1+Y=e^(πs)-e^(2πs)

    (s^2+1)Y=1+e^(πs)-e^(2πs)

    Y(s)=1/(s^2+1) + e^(πs)/(s^2+1) - e^(2πs)/(s^2+1)

    =A + B +C

    2014-10-29 09:23:08 補充:

    £(A)=sin(t)

    £(B)=sin(t-π)δ(t-π)=-sin(t)δ(t-π)

    £(C)=sin(t-2π)δ(t-π)=sin(t)δ(t-π)

    => f(t)=sin(t) + sin(t)[δ(t-π)-δ(t-π)]

    2014-10-29 13:00:21 補充:

    #3.f(t)=minimum{t,1} when 0<=t<∞

    =0(0<=t<1) + 1(t>=1)

    =u(t-1)

    L{u(t-1)}=e^(-s)/s

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  • 5 年前

    #8 的答案應該是 -1 + e^(-t)/2 + (e^t)/2 對嗎 > <

    再請問各位大大 #12,#13 要怎麼解阿QQQQQ

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