求xy (1-x)y=e的n次方的通解-搜答案-智慧作业帮

解x^n=2y^n=3(x²y³)^(2n)-(xy)^3n=(x^n)^4×(y^n)^6-[x^n×y^n]³=2^4×3^6-(2×3)³=16×729-

解x^n=5y^n=3所以（xy)^n=15所以(xy)^2n=[(xy)^n]^2=15^2=225希望对你哟帮助学习进步O(∩_∩)O

(太麻烦拉,给点分啊!)设v=x*x-y*y,u=exp{xy}那么dv/dx=2x(这里应该用偏导符号,代替一下),dv/dy=2y,du/dx=y*exp{xy},du/dy=x*exp{xy}那

原方程是xy=1-e^y?如果是的话将等式两边对X求导数得y+xy'=e^y*y'则y‘=y/(e^y-x)y'(0)=y/e^y

xy-e^x+e^y=0对x求导则(xy)'=1*y+x*y'(e^x)'=e^x(e^y)=e^y*y'所以y-e^x+(x+e^y)y'=0y'=(e^x-y)/(x+e^y)所以dy/dx=(e

x=-2,y=1/2xy=-1x^(2n)*[(xy)^(n+1)]²=(-2)^2n*(-1)^(2n+2)=2^(2n)=4^n

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

（xy）的2n次=225（xy）的n次,当n为奇数时,（xy）的n次为?,所以y的n次=?br/>当n为偶数时,（xy）的n次为15,所以y的n次=3

z=arctan(x*e^x)z'={1/[1+(x*e^x)^2]}*(x*e^x)'(x*e^x)'=x'*e^x+x*(e^x)'=e^x+x*e^x=(x+1)*e^x所以dz/dx=(x+1

(xy)^2n=x^n^2*Y^n^2=5^2*3^2=225

（m+n)x^ny^(m-2)（3xy^2+5x^2y)=21x^my^(n+1)+35x^(m+1)y^n=7x^(m-1)y^(n-1)(3xy^2+5x^2y)m+n=7n=m-1m-2=n-1

xy=e^x-e^yd(xy)=d(e^x-e^y)xdy+ydx=e^xdx-e^ydy(x+e^y)dy=(e^x-y)dx则由dy/dx=(e^x-y)/(e^y+x)

y=e^(x+1);y^n=e^n(x+1)(x→1)lim(x^3-2x+1)/(X^2-1)=1∫(1+xe^5x)/xdx=∫1/xdx+∫e^(5x)dx=lnx+(1/5)e^5x+C

(xy)'=(e^（x+y）'y+xy'=e^(x+y)*(1+y')y'=[e^(x+y)-y]/[1-e^(x+y)]

已知2x+y=4,xy=3,那么2x²y+xy²的值2x²y+xy²=（2x+y）xy =

求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(&#

对x求导y+x*y'=e^(x+y)*(1+y')y+x*y'=e^(x+y)+e^(x+y)*y'所以dy/dx=[e^(x+y)-y]/[x-e^(x+y)]

x^2n=2y^4n=3(xy)^8n=x^8n*y^8n=(x^2n)^4*(y^4n)^2=2^4*3^2=16*9=1444^x*32^y=(2^2)^x*(2^5)^y=2^2x*2^5y=2

xy=e^x-e^y两边求导得：y+xy'=e^x-y'*e^y解得：y'=(e^x-y)/(e^y+x)

min是指f(x)g(x)h(x)三个函数中的最小值