Question

高数题来个详细过程

Answer 1

1. y'' - y' = 1, 特征方程 r^2-r = 0, r = 0, 1. 特解应设为 y = ax， 代入微分方程得 a = -1，则原微分方程的通解是 y = C1+C2e^x -ax. 2. z = xy, (1) 记 F = xy-z， 则 Fx = y, Fy = x, Fz = 1 在点M(1, 1, 1), Fx = 1, Fy = 1, Fz = 1 切平面方程 1(x-1)+1(y-1)+1(z-1) = 0, 即 x+y+z = 3；法线方程 (x-1)/1 = (y-1)/1 = (z-1)/1, 即 x = y = z. (2) 构造拉格朗日函数 F = x^2y^2z^2 - k(xy-z) 则 Fx = 2xy^2z^2-ky = 0 Fy = 2yx^2z^2-kx = 0 Fz = 2zx^2y^2+k = 0 Fk = z-xy = 0 显然， x = y = z = 0 满足 ， 此时 f = x^2y^2z^2 取最小值 3. (1) 积分域 D = D1 + D2， D1 是以点 O(0, 0), A(0, -1), B(1, 0) 为顶点的直角三角形， D2 由 x = √(1-y) 即抛物线 y = 1-x^2 与坐标轴围成。交换积分次序，得 I = ∫dx∫(x+y+1)dy = ∫dx[(x+1)y+y^2/2] = ∫(1/2)(x^4-2x^3-7x^2+4x+4)dx = (1/2)[x^5/5-x^4/2-7x^3/3+2x^2+4x] = 101/60 (2) 积分域关于 坐标平面 yOz, xOz 都对称， x ， y 的奇函数积分为 0. 取柱坐标, I = ∫∫∫zdv = ∫dz∫dt∫zrdr = 2π∫dz[zr^2/2] = π∫z^3dz = π/4

Answer 2

Answer 3

如图所示

Answer 4

两边同时求导可以解得f(x)=-1/x^2
再把f(x)积回去就得到A