微分方程x^2y`+xy= y^2, y|x=1=1 的特解为A x^2y`+xy= y^2, y|x=1=1 B x^2y`+xy= y^2, y|x=1=1
1.函数 (x+y,xy)=(x)^2+(y)^2-xy, 则 f(x,y)=
证明u(x,y)=(x-y)(x^2+4 xy+y^2)u(x,y)=(x-y)(x^2+4 xy+y^2)u(x,y)=(x-y)(x^2+4 xy+y^2)
2.已知 dfrac (x)(x+y)=dfrac (1)(3) ,求 dfrac ({x)^2-(y)^2}(2xy+{y)^2} 的值.
, leqslant xleqslant 2, leqslant yleqslant 2,-|||-其他.-|||-E(X),E(Y),Cov(X,Y),ρ x
()-|||-A (x)=(x)^2-(y)^2+2xy+i((y)^2-(x)^2)-|||-B (x)=(x)^2-(y)^2+2xy+i((y)^2+2x
已知D(X)=4，D(Y)=1，rho xy=0.6，求D(X+Y)，D(3X-2Y)已知D(X)=4，D(Y)=1，，求D(X+Y)，D(3X-2Y)
2.设 (xy,x+y)=(x)^2+(y)^2+xy (其中, =xy =x+y), 则 dfrac (partial f)(partial u)+dfrac
已知z=2xy+(x)/(y)，则当x=1，y=1，△x=1，△y=2时，dz=A. 5B. 6C. 3D. 4
7.设 (x,y)=dfrac (1)(xy),r=sqrt ({x)^2+(y)^2} _(1)= (x,y)|(x,y)in {R)^2 dfrac (1)