Anonymous
Anonymous asked in 科學數學 · 6 years ago

工程數學 Laplace

(一)For reactions-in-series A→B→C,the rate equations for A,B,C can be shown as follows: (k1為A→B之rate k2為B→C之rate)

dA/dt=-k1A

dB/dt=k1A-k2B

dC/dt=k2B

Use Laplace transform to evaluate the following:

1.

(a) Determine A(t),B(t),C(t),if ICs are A(t=0)=1 mol/L B(t=0)=0 C(t=0)=0.

(b) Evaluate the maximal concentration of B(t) and time to achieve this maximum.

2. Change ICs to be A(t=0)=0.7 B(t=0)=0.3 C(t=0)=0 and repeat 1 for all solutions.

(二)假設所有參數均為定值,而非時間函數條件下,推論人口數量化之微分方程表如下:

dP/dt = AP(1-PB^-1) P(0)=x

P:全國人口數 t:時間 A:人口比成長速率 B^-1 :人口最大容納數

(a) 依上述條件求解

(b) dP/dt = AP(1-PB^-1)-kP P(0)=x k:提取出征人口速率 求解

PS 均使用Laplace求 ODEs之解

2 Answers

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  • 麻辣
    Lv 7
    6 years ago
    Favorite Answer

    1.For reactions-in-series x→y→z,the rate equations for x,y,z can be shown as follows: (a=rate for x→y, b=rate for y→z)Ans:£=L^(-1). Change A,B,C into x,y,z for Laplace forms. Change k1, k2 into a, b for simple computation.dx/dt=-a*x........(1)dy/dt=a*x-b*y.....(2)dz/dt=b*y.........(3)Use Laplace transform to evaluate the following:

    (1a) Determine x(t),y(t),z(t), if x(0)=1(mol/L), y(0)=z(0)=0.

    L{Eq.(1)}: s*X(s)-x(0)=-a*X(s) => X(s)=1/(s+a)x(t)=£[1/(s+a)]=e^(-at).....ans

    L{Eq.(2)}: s*Y(s)-0=a*X(s)-b*Y(s) => Y(s)=aX(s)/(s+b)=a/(s+a)(s+b)={1/(s+b)-1/(s+a)}[a/(a-b)]

    y(t)=£{Y(s)}=[a/(a-b)]*[e^(-bt)-e^(-at)].....ans

    L{Eq.(3)}: s*Z(s)-0=b*Y(s) => Z(s)=bY(s)/s=ab/s(s+a)(s+b)=1/s+[b/(s+a)-a/(s+b)]/(a-b)

    z(t)=£{Z(s)}=t+[b*e^(-at)-a*e^(-bt)]/(a-b).....ans

    (1b) Evaluate the maximal concentration of y(t) and time to achieve this maximum.y(t)=[a/(a-b)]*[e^(-bt)-e^(-at)]y'(t)=[a/(a-b)]*[(a/e^at)-(b/e^bt)]=0=> a*e^bt=b*e^at=> a/b=e^(a-b)t=> tm=ln(a/b)/(a-b).....ansmax=y(tm)=[a/(a-b)]*[e^(-b*tm)-e^(-a*tm)]=[a/(a-b)]*(1/e^b-1/e^a)^tm=[a/(a-b)]*(1/e^b-1/e^a)^[ln(a/b)/(a-b)].....ans

    (2) Change ICs to be x(0)=0.7, y(0)=0.3, z(0)=0 and repeat 1 for all solutions.Ans:Please do it by yourself for limitation under 2000 words.

    2.假設所有參數=定值,而非時間函數條件下,推論人口數量化之微分方程表如下:PS 均使用Laplace求 ODEs之解Ans:修改符號適合拉氏轉換格式dy/dt=a*y(1-y/b), y(0)=cy=全國人口數 t=時間 a=人口比成長速率 b=人口最大容納數(a) 依上述條件求解ans:y'=ay-hy^2.....h=a/b=1階2次微分方程式=高次方無法使用拉氏轉變=只有使用柏努力方法來作

    2014-02-28 08:34:25 補充:

    如果可以使用柏努力法

    我會繼續演算

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  • Anonymous
    6 years ago

    到下面的網址看看吧

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

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