小英 發問時間: 科學數學 · 6 年前

工程數學的幾個問題

1.solve y'' + 4y' + 4y = 4cos x +3sin x , y(0)=1 , y'(0)=0

2.given a surface z=e^3x*sin3y , find the plane tangent to the surface at point

(0,π/12,3) .

3.find the work done by the force p = 4xy i - 8y j + 2 k along the straight line y=2x ,

z=2x from point (0,0,0) to point (3,6,6)

4.let y=y(x). solve (e^y + 1)^2*e^-y*dx + (e^x + 1)^2*e^-x*dy = 0 , y(0)=0

5.Evaluate ∫ c (3y - 2x^2) dy + (5x^2 - 3y^2) dx , where C consists of the boundary of the region in the first quadrant that is bounded by the graphs of y = x & y = x^3.

不好意思麻煩了!

1 個解答

評分
  • 麻辣
    Lv 7
    6 年前
    最佳解答

    1.Solve y" + 4y' + 4y = 4cos x + 3sin x, y(0)=1 , y'(0)=0Laplace Transform: (Y*s^2-s)+(Y*s-1)+4Y=4s/(s^2+1) + 3/(s^2+1)Y(s^2+s+1)=(4s+3)/(s^2+1) + (s+1)Y=[(4s+3)+(s+1)(s^2+1)]/(s^2+1)(s^2+s+1)=(s^3+s^2+5s+4)/(s^2+1)(s^2+s+1)=(as+b)/(s^2+1) + (cs+d)/(s^2+s+1)f(s)=s^3+s^2+5s+4=(as+b)(s^2+s+1) + (cs+d)(s^2+1)=(a+c)s^3+(a+b+d)s^2+(a+b+c)s+(b+d)a+c=1, b+d=41=a+b+d=a+4 => a=-35=a+b+c=b+1 => b=4=> c=4, d=0Y(s)=(-3s+4)/(s^2+1) + 4s/(s^2+s+1)=-3s/(s^2+1)+4/(s^2+1) + 4[(s+1/2)-1/2]/[(s+1/2)^2+3/4]=-3s/(s^2+1)+4/(s^2+1)+4(s+1/2)/[(s+1/2)^2+3/4]-2/[(s+1/2)^2+3/4]y(t)=-3cos(t)+4sin(t)+4e^(-t/2)cos(√3t/2)-(4/√3)e^(-t/2)sin(√3t/2)

    2.Given a surface z=e^3x*sin3y, find the plane tangent to the surface at point (0,π/12,3).N= Normal= ▽(e^3x*sin3y - z)= 3e^3x*sin3y i + 3*e^3x*cos3y j - k= 3*1*sin(π/4) i + 3*1*cos(π/4) j - k= 3√2/2 i + 3√2/2 j - k=> Tangent plane: 0 = (3√2/2)*x + (3√2/2)(y-π/12) - (z-3)

    3.find the work done by the force p = 4xy i - 8y j + 2 k along the straight line y=2x, z=2x from point (0,0,0) to point (3,6,6)y=z=2x => dy=dz=2dxw = ∫p.dL= (4xy i - 8y j + 2 k).(dx i + dy j+ dz k)=∫(4xydx - 8ydy + 2kdz)=∫(4x*2xdx - 8*2x*2dx + 2k*2dx)=∫<0~3>(8x^2 - 32x + 4)dx=8x^3/3 - 16x^2 + 4x=8*27/3 - 16*9 + 4*3=72 - 144 + 12=-60

    4.Solve (e^y+1)^2*e^(-y)*dx+(e^x+1)^2*e^(-x)*dy=0, y(0)=00=dx/(e^x+1)^2*e^(-x) + dy/(e^y+1)^2*e^(-y)=e^x*dx/(e^x+1)^2 + e^y*dy/(e^y+1)^2c=∫d(e^x+1)/(e^x+1)^2 + ∫d(e^y+1)/(e^y+1)^2=-1/(e^x+1) - 1/(e^y+1)=-1/2 - 1/2=-1=> 0 = 1 -1/(e^x+1) - 1/(e^y+1)

    5.Evaluate ∫C (3y-2x^2)dy - (-5x^2+3y^2)dx, where C consists of the boundary of the region in the first quadrant that is bounded by the graphs of y = x & y = x^3.∫-(-5x^2+3y^2)dx + (3y-2x^2)dy =∫-N*dx + M*dy=∫∫(Mx + Ny)dA......Green's Threom=∫∫(-4x + 6y)dydx=∫(-4xy + 3y^2)dx.....y=x~x^3 =∫[-4x(x^3-x) + 3(x^6-x^2)]dx=∫(3x^6 - 4x^3 + x^2)dx=3x^7/7 - x^4 + x^3/3......x=0~1=3*1/7 - 1 + 1/3=6/7 -21/21 + 2/3=11/21

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