請教這兩題工程數學。

1.y'=(6xy-y^2)/(3xy-6x^2)Ans:dy/dx=y(6x-y)/3x(y-2x).....1階齊次方程式0=y(6x-y)dx+3x(2x-y)dy.....令y=ux=ux(6x-ux)dx+3x(2x-ux)(udx+xdu).....dy=udx+xdu=u(6x-ux)dx+3(2x-ux)(udx+xdu)=u(6-u)dx+3(2-u)(udx+xdu)=u(6-u)dx+3(2-u)udx+3(2-u)xdu=[u(6-u)+3u(2-u)]dx+3(2-u)xdx=u(6-u+6-3u)dx+3(2-u)xdx=u(12-4u)dx+3(2-u)xdu=dx/x+3(u-2)du/4u(u-3)ln(c1)=∫dx/x+∫du/4(u-3)-∫du/4u=ln(x)+0.25*ln(u-3)-0.25*ln(u)=ln{x*[(u-3)/u]^0.25}c1=x[(u-3)/u]^0.25c1^4=[(u-3)/u]*x^4u=c*(u-3)*x^4.....c=1/c1^4y/x=c*[(y/x)-3)]*x^4y=c*(y-3x)*x^4=c*y*x^4-3c*x^5y(cx^4-1)=3cx^5y(x)=3cx^5/(cx^4-1).....ans

2014-02-13 15:04:34 補充：
(3) 逆運算子部份分式法

yp(x)=(aD+b)/(D^2+9)+c/(D+2)+f/(D-2)

f(D)=1=(D^2-4)(aD+b)+c(D-2)(D^2+9)+f(D+2)(D^2+9)

f(2)=1=f*4*13 => f=1/52

f(-2)=1=-4*52c => c=-1/52

f(0)=1=-4b-2*9c+2*9f => b=-1/13

2014-02-13 15:05:07 補充：
f(1)=1=(a+b)(-3)+c(-1)10+f*3*10 =>

1=-3*(a-1/13)+10/52+30/52

=-3a+3/13+40/52

=-3a+3/13+10/13

=-3a+13/13

=-3a+1 => a=0

yp(x)=-1/13(D+9^2)-1/52(D+2)+1/52(D-2).....Note

2014-02-13 15:06:31 補充：
(3.1) D=2 => y=e^(2x)

y1(x)=[e^(2x)/52]{∫e^(-2x)10*e^(2x)dx+∫3e^(-2x)*cos3x*dx}.....Note1

=[e^(2x)/52]{10x+3[a*cos(bx)+b*sin(bx)]e^(ax)/(a^2+b^2)}.....a=-2,b=3

=[e^(2x)/52]{10x+3[-2cos(3x)+3sin(3x)]e^(-2x)/13}

=(5/26)*x*e^(2x)+(3/676)*[-2cos(3x)+3sin(3x)]

2014-02-13 15:07:05 補充：
Note1: 部份積分法

A=∫e^(ax)*cos(bx)*dx

=∫cos(bx)d(e^ax)/a

a*A=cos(bx)*e^(ax)+b∫e^(ax)*sin(bx)*dx

=cos(bx)*e^(ax)+b∫sin(bx)*d(e^ax)

=cos(bx)*e^(ax)+b[e^ax*sin(bx)-b∫e^ax*cos(bx)dx]/a

2014-02-13 15:07:27 補充：
A*a^2=a*cos(bx)*e^(ax)+b[e^ax*sin(bx)-b*A]

=a*cos(bx)*e^(ax)+b*e^ax*sin(bx)-A*b^2

A(a^2+b^2)=a*cos(bx)*e^(ax)+b*e^ax*sin(bx)

A=[a*cos(bx)+b*sin(bx)]e^(ax)/(a^2+b^2)

2014-02-13 15:07:51 補充：
(3.2) D=-2 => y=e^(-2x)

y2(x)=[-e^(-2x)/52]{∫e^(2x)10*e^(2x)dx+∫3e^(2x)*cos3x*dx}

=[-e^(-2x)/52]{10∫e^(4x)dx+3∫e^(2x)*cos3x*dx}.....a=2,b=3

=[-e^(-2x)/52]{(5/2)*e^(4x)+(3/13)*(2cos3x+3sin3x)e^(2x)}

=-(1/52)*{(5/2)*e^(2x)+(3/13)*(2cos3x+3sin3x)}

=-{(5/104)*e^(2x)+(3/676)*(2cos3x+3sin3x)}

2014-02-13 15:51:08 補充：
有事出去

2014-02-13 18:58:07 補充：
超過字限
請轉意見欄

2014-02-13 18:58:20 補充：
(3.4) D=-3j => y=e^(-3jx)

y3(x)=[-e^(3jx)/13]{∫e^(-3jx)10*e^(2x)dx+∫3e^(-3jx)*cos3x*dx}

=[-e^(-3jx)/13]{10*e^(2-3j)x/(2-3j)-e^(-3jx)*(sin3x-cos3x)/2}

=-10e^(2-3j)/13(2-3j)+(sin3x-jcos3x)/26

2014-02-13 18:59:10 補充：
(3.3) D=+3j => y=e^(3jx)

y3(x)=[-e^(-3jx)/13]{∫e^(3jx)10*e^(2x)dx+∫3e^(3jx)*cos3x*dx}

=[-e^(-3jx)/13]{10*e^(2+3j)x/(2+3j)+e^(3jx)*(sin3x-jcos3x)/2}.....Note2

=-10e^(2+3j)/13(2+3j)-(sin3x-jcos3x)/26

2014-02-13 19:10:59 補充：
Note2: 虛函數部份積分

B=∫e^(ajx)cos(ax)dx

=∫e^(ajx)d(sin(ax))/a

aB=sin(ax)e^(ajx)-aj∫sin(ax)e^(ajx)dx

=e(ajx)sin(ax)-j∫e^(ajx)d(cos(ax))

=e(ajx)sin(ax)-j[e^(ajx)cos(ax)-aj∫cos(ax)e^(ajx)dx]

=e(ajx)sin(ax)-je^(ajx)cos(ax)+aj^2*B

2aB=e(ajx)[sin(ax)-jcos(ax)]

B=e(ajx)[sin(ax)-jcos(ax)]/2a

2014-02-13 19:12:04 補充：
(3.5) 兩者合併

y5=y3+y4

=-10e^(2+3j)/13(2+3j)-(sin3x-jcos3x)/26

-10e^(2-3j)/13(2-3j)+(sin3x-jcos3x)/26

=-10e^(2+3j)/13(2+3j)-e^(2-3j)/13(2-3j)

=-10[(2-3j)e^(2+3j)+(2+3j)e^(2-3j)]/169

=-10e^2[(2-3j)e^3j+(2+3j)e^(-3j)]/169

=-10e^2{(2-3j)(cos3x+jsin3x)+(2+3j)(cos3x-jsin3x)}

2014-02-13 19:12:26 補充：
=-10e^2{2*cos3x-6j^2*sin3x}

=-20e^2*(cos3x+2sin3x)

2014-02-13 19:16:42 補充：
(3.6) 特殊解

yp(x)=y1+y2+y5

=(5/26)*x*e^(2x)+(3/676)*[-2cos(3x)+3sin(3x)]+

-(1/52)*{(5/2)*e^(2x)+(3/13)*(2cos3x+3sin3x)}+

-20e^2*(cos3x+2sin3x)+

2014-02-13 19:17:22 補充：
(3.7) 通解

y(x)=yh(x)+yp(x)

=請自己代入

到下面的網址看看吧

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

1.y'=(6xy-y^2)/(3xy-6x^2)

(Sol)(3xy - 6x^2) dy = (6xy – y^2) dx
(y^2 – 6xy) dx + (3xy – 6x^2) dy = 0
令 M = y^2 -6xy, N = 3xy – 6x^2
∂M/∂y = 2y - 6x, ∂N/∂x = 3y – 12x(∂M/∂y - ∂N/∂x)/N = (2y -6x -3y +12x)/(3xy-6x^2) = (6x –y)/(3xy – 6x^2)
令 I(x,y) = (x^a)(y^b)乘上原式
{(X^a)[y^(b+2)] - 6[x^(a+1)][y^(b+1)]} dx+{3[x^(a+1)][y^(b+1)] – 6[x^(a+2)](y^b)} dy= 0
∂M/∂y = (b+2)(x^a)[y^(b+1)] – 6(b+1)[x^(a+1)](y^b)∂N∂x = 3(a+1)(x^a)[y^(b+1)] –6(a+2)[x^(a+1)](y^b)由 ∂M/∂y = ∂N/∂x得b + 2 = 3a + 3b + 1 = a + 2解得 a = 0, b =1所以 I(x,y) = y 原方程式寫為(y^3 – 6xy^2) dx + [3xy^2 – 6(x^2)y] dy = 0令Φ(x,y) = ∫(y^3 – 6xy^2) dx= xy^3 – 3(x^2)(y^2) + f(y)∂Φ/∂y = 3xy^2 – 6(x^2)y + f ‘(y)比較原方程式得 f ‘(y)= 0f(y) = c故解為 xy^3 –3(x^2)(y^2) + c = 0 2. 這一題麻煩用逆運算子。
(D^4+5D^2-36)y=10(e^-2x)+3cos(3x) (Sol)過程繁雜，只幫你解特解的部份分兩部分，第一部分
(D^4 + 5D^2 - 36)y = 10e^(-2x) y = [1/(D^4 + 5D^2 – 36)] 10e^(-2x)= {1/[(D^2 + 9)(D^2 – 4)]} 10e^(-2x)
= 10e^(-2x) {1/【[(D - 2)^2 +9][(D – 2)^2 – 4]】}*1= 10e^(-2x) {1/[(D^2 – 4D + 13)D(D – 4)]}*1
= 10e^(-2x) {1/[(D^2 – 4D + 13)(D – 4)]}x
= 10e^(-2x) (-1/4){1/[(D^2 – 4D + 13)(1 – D/4)]}x
= (-5/2)e^(-2x) [(1+D/4)/(D^2 – 4D + 13)]x
= (-5/2)e^(-2x) [1/(D^2 – 4D + 13)](x + 1/4)
= (-5/2)e^(-2x) (1/13){1/[1 – 4D/13 + (D^2)/13]}(x + 1/4)
= (-5/26)e^(-2x) (1 + 4D/13)(x + 1/4)= (-5/26)e^(-2x) (x + 1/4 + 4/13)
= (-5/26)e^(-2x) (x + 29/52) ………………………..(1)

字數限制，其他轉意見欄


2014-02-13 19:02:27 補充：
第二部分
(D^4 + 5D^2 - 36)y = 3cos 3x
y = [3/(D^4 + 5D^2 – 36)] cos 3x

= [3/(D^4 + 5D^2 – 36)] [e^(i3x) + e^(-i3x)]/2
= (3/2)[1/(D^4 + 5D^2 – 36)] e^(i3x) + (3/2)[1/(D^4 + 5D^2 – 36)] e^(-i3x)

2014-02-13 19:03:18 補充：
= (3/2){[1/[(D^2 + 9)(D^2 – 4)] e^(i3x) + (3/2){[1/[(D^2 + 9)(D^2 – 4)] e^(-i3x)
= (3/2)e^(i3x){[1/【[(D +3i)^2 + 9][D +3i]^2 – 4)]】}*1
+ (3/2)e^(-i3x){[1/【[(D -3i)^2 + 9][D -3i]^2 – 4)]】}*1

2014-02-13 19:03:43 補充：
= (3/2)e^(i3x){[1/[D(D +6i)(D +2 +3i)(D –2 +3i)]}*1
+ (3/2)e^(-i3x){[1/[D(D -6i)(D +2 -3i)(D –2 -3i)]}*1
= (3/2)e^(i3x){1/【(6i)(2 +3i) (-2 +3i)D[1 +D/(6i)][1+D/(2 +3i)][1 –D/(-2 +3i)]】}*1
+ (3/2)e^(-i3x){1/【(-6i)(2 -3i) (-2 -3i)D[1 -D/(6i)][1+D/(2 -3i)][1 –D/(-2 -3i)]】}*1

2014-02-13 19:03:57 補充：
= (3/2)e^(i3x) [x/(-78i)] + (3/2)e^(-i3x)[x/(78i)]
= xe^(i3x)/(-52i) + xe^(-i3x)/(52i)
= -(x/26){[e^(i3x) – e^(-i3x)]/(2i)}

= (-x/26) sin 3x ………………………………………..(2)



由(1)(2)得

y = (-5/26)e^(-2x) (x + 29/52) +(-x/26) sin 3x