y＇ xy=e的x次方是什么一阶线性方程-搜答案-搜搜做题作业网

设y=C(x)e^(-∫dx)=C(x)e^(-x)代入原微分方程C‘(x)e^(-x)-C(x)e^(-x)+C(x)e^(-x)=e^xC‘(x)e^(-x)=e^xC‘(x)=e^(2x)C(x

y'=1/(x+e^y)x‘=x+e^yx=Ce^y+ye^y再问：你看错了，是这个题目y'=1/x+e^y再答：哦y'=1/x+e^ye^(-y)y'=e^(-y)/x+1xe^(-y)y'=e^(

(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

根据一阶全微分形式不变得dz=d(xf(x^y,e^xy)=f(x^y,e^xy)dx+xd(f(x^y,e^xy))=f(x^y,e^xy)dx+x[f1'd(x^y)+f2'(de^xy)]=f(

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

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+(&#

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=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)

你分别算就行了,e的y次方求导,就是e的y次方乘以y的导数,再算xy,求导等于y+x和y的导数的乘积,最后把这两个结果夹在一起就行了再问：哦

1.(1)au/ax=f1'*(x^2-y^2)'x+f2'*(e^xy)'x=2x*f1'+y(e^xy)*f2'其中,f'1表示对第一个变量求偏导数(x^2-y^2)'x表示对x求偏导数au/ay

令u=xy,v=e^(x+y)Z'x=Z'u*U'x+Z'v*V'x=f'u*y+f'v*e^(x+y)Z'y=Z'u*U'y+Z'v*V'y=f'u*x+f'v*e^(x+y)

令a=x^2-y^2b=e^(xy)f具有一阶连续偏导数f1‘和f2’∂u/∂x=(∂u/∂a)×(∂a/∂x)+(∂

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

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)]

x2+1)(y2-1)dx+xydy=0ydy/(y^2-1)=-(x+1/x)dx两边积分(1/2)ln|y^2-1|=-x^2/2-ln|x|+C1ln|y^2-1|=-x^2-2ln|x|+2C

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)