(x∧2 1)dy dx+2xy=4x∧2

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(x∧2 1)dy dx+2xy=4x∧2
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)

1.x-z/xy - 2ab/xy

1=(x-z-2ab)/xy2=(a²-2ab+b²)/a-b=(a-b)²/a-b=a-

(xy-x^2)乘以(xy)/(x-y)

对.前提是x不等于y

(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=

(3xy)平方/(-xy)+(x-2y)平方-(x+2y)(x-2y)=-9xy+(x-2y)(x-2y-x-2y)=-9xy+(x-2y)(-4y)=-9xy-4xy+8y²=8y

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

若正实数x ,y满足2x+y+6=xy.则xy的最小值.

2x+y+6≥6+2√2xyxy≥6+2√2xy(√xy-√2)^2≥8√xy-√2≥2√2或√xy-√2≤-2√2(不可能)所以xy最小值是(3√2)^2=18-------------------

x∧2y +xy∧2-e∧xy=0所确定的函数y=y(x)的导数 急死了

x^2y+xy^2-e^xy=0对x求导得2xy+x^2y'+y^2+x*2yy'-(ye^xy+e^xy*y')=02xy+y^2-ye^xy+(x^2+2xy-e^xy)y'=0(x^2+2xy-

(1+x∧2)dy=(1+xy)dx,y(x=1)=0求特解

(1+x∧2)dy=(1+xy)dx,y(x=1)=0观察知,y=x是方程的特解为求通解,令y=x+t,代入原方程得(1+x^2)(1+t')dx=(1+x^2+xt)dx化简得dt/t=xdx/(1

(x^2+xy/x-y)/(xy/x-y)计算

【x²+xy/(x-y)】/【xy/(x-y)】=【x²(x-y)/(x-y)+xy/(x-y)】/【xy/(x-y)】={【x²(x-y)+xy】/(x-y)}/【xy

已知(x+2)∧2+y+1的绝对值=0,求5xy∧2-{2x∧2y-【3xy^2】-(4xy^2-2x∧2y)}的值

根据已知得到x=-2y=-1原式=5xy^2-{2x^2y-[3xy^2-(4xy^2-2x^2y)]}=5xy^2-{2x^2y-[3xy^2-4xy^2+2x^2y]}=5xy^2-{2x^2y-

(-1/3xy)∧2*[3xy(2x-y)-3x(2xy-4y∧2)]

(-1/3xy)∧2*[3xy(2x-y)-3x(2xy-4y∧2)]解得:1/9x*x*y*y*[3xy(2x-y)-6xy(x-2y)]==(1/9x*x*y*y*)*3xy*[2x-y-2(x+

-3(2x^2-xy)+4(x^2+xy-6)

-3(2x^2-xy)+4(x^2+xy-6)=-6x^2+3xy+4x^2+4xy-24=-2x^2+7xy-24再问:thankyou~

求微分方程dydx+y=e

这是一阶线性微分方程,其中P(x)=1,Q(x)=e-x∴通解y=e−∫dx(∫e−x•e∫dxdx+C)=e−x(∫e−x•exdx+C)=e−x(x+C).

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

因式分解:-xy^2+2xy-x

-xy^2+2xy-x=-x(y^2-2y+1)=-x(y-1)^2如果本题有什么不明白可以追问,如果满意请点击“选为满意答案”

分解因式:(X+Y-2XY)*(X+Y-2)+(1-XY)∧2

你第一个是不是写错了?[2][(x-1)(y-1)]^2[3](x^3-x+1)(x^2+x+1)

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.

x-y/x²÷x²-2xy+y²/xy*(xy-x²)

(x-y)/x²÷(x²-2xy+y²)/xy×(xy-x²)=(x-y)/x²÷(x+y)²/xy×x(y-x)=(x-y)²/

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).