y(2y-y)=y^2*sinx

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1.5y+8y-6y=2y-3

移项、得1.5y=-3系数化为一、y=-2很简单的呀

证明COS(X+Y)COS(X-Y)=COS^2X-SIN^2Y

COS(X+Y)COS(X-Y)=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN

tan(x+y)tan(x-y)=sin^2x-sin^2y/cos^2x-sin^2y 顺便问一下. tan,sin,

tan,正切;sin,正弦;cos,余弦tan(x+y)tan(x-y)=sin(x+y)/cos(x+y)*sin(x-y)/cos(x-y)=sin(x+y)sin(x-y)/[cos(x+y)c

一道三角恒等式证明题请证明sin(x+y)sin(x-y)=sin^2(x)-sin^2(y)

左边=(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=sin²xcos²y-cos²xsin²y=sin²x(1-sin

y =(cos^2) x - sin (3^x),求y'

y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos

证明sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((x+y)/2)sin((x+

sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/

sin(x+y)sin(x-y)=k,求cos^2x-cos^2y

-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k

求微分方程y'+sin[(x+y)/2]=sin[(x-y)/2]通解

(1)当y=C时,sin[(x+C)/2]=sin[(x-C)/2]移项,和差化积有2cos{[(x+C)/2+(x-C)/2]/2}sin{[(x+C)/2-(x-C)/2]/2}=0,即cos(x

求导:x^2*y^2 + x sin(y) = 1

对这样的隐函数求导数的时候,就把y看作x的函数,y对x求导就得到dy/dx所以原等式对x求导得到2xy²+x²*2y*dy/dx+siny+x*cosy*dy/dx=0于是化简得到

2y''+y'-y=0

2y''+y'-y=0特征方程:2r^2+r-1=0根为:-1,1/2y=C1e^(-x)+C2e^(x/2)

y=sin(2x+30度)求导,y'=?

过程:先将括号里的当作一个整体,即求sinx的导数,所以是cos(2x+30度),再对括号里的求导,所以得2由复合函数的求导法则,知y=2cos(2x+30度)

函数y=sin(π2

y=sin(π2+x)cos(π6-x)=cosx(32cosx+12snx)=32cos2x+12sinxcosx=34(1+cos2x)+14sin2x=12sin(2x+π3)+34∴T=2π2

求证1、sin l-sin y=2cos(l+y)/2 sin(l-y)/2

前三题其实就是和差化积的公式,4因为tan2a=2tana/(1-tan^2a)sin2a=2tana/(1+tan^2a)所以左边=2tana/(1+tan^2a)-√3cos2a.先消去一个tan

求y=sin(e^2x)的微分y'

y'=2e^2xcos(e^2x)把y看成复合函数sint,t=e^m,m=2x.复合函数求导,等于三个分别求导的积

化简sin(2x-y)*sin y+cos(2x-y)*sin y

我来给你解答,稍等再答:

lim sin(y×x^2+y^4)/(x^2+y^2) x,y都趋于0,

令y=kx则limsin(y×x^2+y^4)/(x^2+y^2)=limsin[kx^3+(kx)^4]/[(1+k^2)*x^2]分子用等价无穷小替换=lim[k+(k^4)*x]*(x^3)/[