dy dx=(-2x y)^2-7

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/19 01:24:23
求由方程xy=ex+y所确定的隐函数的导数dydx

方程两边求关x的导数ddx(xy)=(y+xdydx);     ddxex+y=ex+y(1+dydx);所以有  (y+xdy

求解微分方程dydx

由微分方程dydx=2xy,得dyy=2xdx(y≠0)两边积分得:ln|y|=x2+C1即y=Cex2(C为任意常数)

先化简,再求值 ⒈2(Xy+Xy)-3(Xy-xy)-4Xy,其中X=1,y=-1

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

已知A=5x²-3xy²+2xy,B=7xy²-2xy+x²y

A-2B=5x²-3xy²+2xy-2(7xy²-2xy+x²y)=5x²-3xy²+2xy-14xy²+4xy-2x²

已知x2-xy=7,2xy+y2=4,则代数式x2+xy+y2的值是___.

∵x2-xy=7,2xy+y2=4,∴原式=(x2-xy)+(2xy+y2)=7+4=11,故答案为:11

设函数y=y(x)由方程ln(x2+y)=x3y+sinx确定,则dydx|

方程两边对x求导得2x+y′x2+y=3x2y+x3y′+cosxy′=2x−(x2+y)(3x2y+cosx)x5+x3y−1由原方程知,x=0时y=1,代入上式得y′|x=0=dydx|x=0=1

已知x+y=4,xy=2,求代数式2(3x+2xy)+(2xy+3x+2y)-(4xy-7y)的值

原式=6x+4xy+2xy+3x+2y-4xy+7y=9x+9y+2xy=9(x+y)+2xy=9×4+2×2=36+4=40

当x=3,y=3分之1时,求代数出3xy-[2xy-2(xy-2分之3xy)+xy]+3xy的值

3xy-[2xy-2(xy-2分之3xy)+xy]+3xy=6xy-[2xy-2xy+3xy+xy)=6xy-4xy=2xy=2×3×3分之1=2

已知x2+xy=-3,xy+y2=7,试求:x2+2xy+y2的值.

∵x2+xy=-3,xy+y2=7,∴(x2+xy)+(xy+y2)=-3+7=4,即x2+2xy+y2=4.

xy=-2,x+y=3,求代数式(4xy+12y)+[7x-(3xy+4y-x)]

(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22

已知xy^2=-2求xy(2x^3y^7+5x^2y

最简单的方法就是用特殊值,令x=-2,y=1然后代入所求表达式,求出其值.为-8;另外此内题的另一个解法是变化所求表达式,使它变成关于xy^2的表达式.

已知a=5x的平方y-3xy的平方+4xy,b=7xy的平方-2xy+x的平方y.

a-2b=(5x²y-3xy²+4xy)-2(7xy²-2xy+x²y)=5x²y-3xy²+4xy-14xy²+4xy-2x&#

化简:xy分之3x^2+2xy-xy分之2x^2-xy=

(3x^2+2xy)/xy-(2x^2-xy)/xy=(3x^2+2xy-2x^2+xy)/xy=(x^2+3xy)/xy=x(x+3y)/xy=(x+3y)/y

x+y=8,xy=-2,求(5xy+4x+7y)+(6x-3xy)-(4xy-3y)的值

(5xy+4x+7y)+(6x-3xy)-(4xy-3y)=5xy+4x+7y+6x-3xy-4xy+3y=(5xy-3xy-4xy)+(4x+6x)+(7y+3y)=-2xy+10x+10y=-2x

3xy-3xy-xy+2yx

3xy-3xy-xy+2yx=-xy+2xy=xy

matlab solve函数 xmaxr=solve(dydx,x)

dydx要是等式才行吧.如果是的话,这句话就是求这个等式的根,用r表示x.

已知A=8x^2y-6xy^2-3xy,B=7xy^2-2xy+5x^2y,

∵A=8x²y-6xy²-3xy,B=7xy²-2xy+5x²y ∴C=3(A+B)  =3(8x²y-6xy²-3xy+7xy²

已知A=5x^2y-3xy^2+4xy,B=7xy^2-2xy+x^2y.

已知A=5x^2y-3xy^2+4xy,B=7xy^2-2xy+x^2y.(1)求A-2B的值A-2B=5x^2y-3xy^2+4xy-2*(7xy^2-2xy+x^2y)=5x^2y-3xy^2+4

已知x²-7xy+12y²=0求x²-xy+y²/2xy的值

x²-7xy+12y²=0(x-3y)(x-4y)=0x1=3yx2=4yx=3y时原式=9y²-3y²+y²/6y²=7/6x=4y时原式

设函数y=y(x)由方程ex+y+cos(xy)=0确定,则dydx

在方程ex+y+cos(xy)=0左右两边同时对x求导,得:ex+y(1+y′)-sin(xy)•(y+xy′)=0,化简求得:y′=dydx=ysin(xy)−ex+yex+y−xsin(xy).