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一年打兩次瘦小臉,50年不就要打100次? 不用!彥安診所:定形式瘦小臉不再肉餅臉

工程數學 (急) 20點

1. 5y"+7y’+2y=2e^2x

特徵方程式為:

0=5D^2+7D^2+2=(D+1)(5D+2)

D=-1,-2/5

=> yh = a*e^(-x) + b*e^(-2x/5)

yp = ce^2x

yp’= 2ce^2x

yp"= 4ce^2x

代入原式裡面:

2e^2x=(20+14+2)ce^2x

=> c=1/18

=> yp=e^2x/18

y= yh+yp

= a*e^(-x) + b*e^(-2x/5) + e^2x/18


2. 5y"+7y’+2y=4sin2x

Similar to Problem-1

yh = a*e^(-x) + b*e^(-2x/5)

yp = c*cos2x + d*sin2x

yp’= -2c*sin2x + 2d*cos2x

yp"= -4c*cos2x – 4d*sin2x

代入原式裡面:

4*sin2x=-20c*cos2x-20d*sin2x-14c*sin2x+14d*cos2x+2c*cos2x+2d*sin2x

=(-20c+14d+2c)cos2x + (-20d-14c+2d)sin2x

=(-18c+14d)cos2x + (-14c-18d)sin2x

=> 7c+9d=-2, 9c=7d

=> c=-7/65, d=-9/65

=> yp = -(7*cos2x + 9*sin2x)/65

=> y = a*e^(-x) + b*e^(-2x/5) – (7*cos2x + 9*sin2x)/65

3. 5y"+7y’+2y=9

yp=9/2 by inspection

=> y = a*e^(-x) + b*e^(-2x/5) + 9/2

4. 5y"+7y’+2y=9+4sin2x+2e^2x

yp’s can be sum up due to linear equation

=> yp = e^2x/18 – (7*cos2x + 9*sin2x)/65 + 9/2

=> y=a*e^(-x)+b*e^(-2x/5)+e^2x/18-(7*cos2x+9*sin2x)/65-9/2

5. 4y"+y=0

0=4D^2+1 => D=+-j/2, j=image

y= a*cos(x/2) + b*sin(x/2)


6. y"+4y’+4y=0

0 = D^2 + 4D + 4 = (D+2)^2

D = -2, -2

y = (a + bx)*e^(-2x)

7. 2y"-3y’+4y=0

0 = 2D^2 – 3D + 4

D = 3/4 +- √23j/4

y = e^(3/4)*[a*cos(kx) + b*sin(kx)]…..k=√23/4

8. 某微分方程式特徵值λ=4,-1,2,2,2,2,共6根 試寫出其齊次解(自變數可自設)

y = c1*e^4x + c2*e^(-x) + (c3 + c4*x + c5*x^2 + c6*x^3)*e^2x

9. 5y"+7y’+2y=2e^2x, y(0)=y’(0)=1

y(x) = a*e^(-x) + b*e^(-2x/5) + (1/18)*e^2x

y’(x) = -a*e^(-x) – (2b/5)*e^(-2x/5) + (1/9)*e^2x

y(0) = a + b + 1/18 = 1 => 18a + 18b = 17

y’(0) = – a – 2b/5 + 1/9 = 1 => 45a + 18b + 40 = 0

=> a = -19/9,  b = 55/18

=> y(x) = (-19/9)*e^(-x) + (55/18)*e^(-2x/5) + (1/18)*e^2x

10.試說明特徵方程式與特徵值所代表的物理意義

m=mass of car, c=damping, k=spring constant

m*x" + c*x’ + k*x = F*sin(wt)

特徵方程式: 0 = m*D^2 + c*D + k => 用來判斷系統的振動狀況

1.虛根 = 上下振動非常厲害不會停止

2.實根 = 變成避震器馬上停止

1x1.trans 工程數學 (急) 20點
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