dy dx=sin(y)

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dy dx=sin(y)
求y=sin(x+1)周期

y=Asin(ωx+ψ)周期为:T=2π/ωy=sin(x+1),ω=1,所以T=2π

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)

证明sin(x+y)sin(x-y)=sinx-siny

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2(sinx)^2-1+2(siny)^2)=(si

y=sin(sinx)求导

y=sin(sinx)y‘=cos(sinx)*(sinx)'=cos(sinx)*cosx

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

求证 cos*xcos*y + sin*xsin*y + sin*xcos*y + xin*ycos*x = 1

合并同类项么,很简单的只要你愿意去做左边=cos*x(cos*y+sin*y)+sin*x(cos*y+sin*y)=cos*x+sin*x=1=右边

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=sin[sin(x^2)] 则dy/dx=?

dy/dx相当于对x进行求导:dy/dx=y'=2x*cos[sin(x^2)]*cos(x^2)由于:sinx=cosx,sin(x^2)=2x*cos(x^2)

x*y'*sin(y/x)-y*sin(y/x)+x=0 求微分方程的解

y'sin(y/x)-y/x*sin(y/x)+1=0令y/x=u,则y'=u+xu'所以(u+xu')sinu-usinu+1=0xu'sinu+1=0-sinudu=dx/x两边积分:cosu=l

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)/

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).

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

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.

求证:sin(x-y)sin(x+y)/sin²xcos²y=1-cot²xtan

sin(x-y)=sinxcosy-cosxsiny,sin(x+y)=sinxcosy+cosxsinysin(x-y)sin(x+y)=sin²xcos²y-cos²

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

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).