y导数等于e的x y次方的通解
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e的x次方对x求导还是e的x次方乘以dy/dxxy是复合函数需要分别求导先x求导是y然后y求导是x乘以dy/dx这是用复合函数求导公式得来的-e对x求导是0至于为什么有的有dy/dx而y那项没有dy/
直接求导(xy^2)=y^2+2xy*y'(e^xy)'=(xy)'*xy*e^xy=(y+x*y')*xy*e^xy然后带进去求y'就是dy/dx
对x求导为y*e^(xy)对y求导为x*e^(xy)对x,y求偏导为e^(xy)+xy*e^(xy)
y'=(x)'(e^(-2x))+(x)(e^(-2x))'=e^(-2x)+x(e^(-2x))(-2)=(e^(-2x))*(1-2x)
y²=e^(x+y)两边对x求导2yy'=e^(x+y)*(1+y')2yy'-e^(x+y)*y'=e^(x+y)y'=e^(x+y)/(2y-e^(x+y))y'=y²/(2y
e^y-xy=ee^y·dy/dx-(y+x·dy/dx)=0e^y·dy/dx-y-x·dy/dx=0(e^y-x)·dy/dx=ydy/dx=y/(e^y-x)dy/dx不能叫做dx分之dy,因为
e^(x+y)-e^x+[e^(x+y)+e^y]•dy/dx=0[e^(x+y)+e^y]•dy/dx=e^x-e^(x+y)=e^x•(1-e^y)dy/dx=
该题为隐函数求导.xy+e^(xy)=1则y+xy'+e^(xy)(y+xy')=0解得:y'=-y/x解答完毕.
题目应该是y"+3y'+2y=e^x吧?特征方程为r^2+3r+2=0,得r=-1,-2即齐次方程的通解y1=C1e^(-x)+C2e^(-2x)设特解y*=ae^x,代入方程得:ae^x+3ae^x
直接运用公式y=x^2+e^x,有y'=2x+e^x
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+x*y'=e^(x+y)*(1+y')∴dy/dx=[e^(x+y)-y]/[x-e^(x+y)].
(xy)'=(e^(x+y)'y+xy'=e^(x+y)*(1+y')y'=[e^(x+y)-y]/[1-e^(x+y)]
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)]
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)
(1+x^2)e^yy'-2x(1+e^y)=0令u=1+e^y则u'=e^y*y'代入方程得:(1+x^2)u'-2xu=0因此有:du/u=2xdx/(1+x^2)即:du/u=d(x^2)/(1