ysinx=ylny,x= ,y=e

（一题）从这步d(ysinx)-dcos(x-y)=0到这步sinxdy+ycosxdx+sin(x-y)(dx-dy)=0不懂是么?ysinx是两个数相乘,对它d(ysinx)时就得用公式d(UV)

ysinx=cos(xy)两边分别求导y'sinx+ycosx=-sin(xy)(y+xy')y'=-y(sin(xy)+cosx)/(sinx+xsin(xy))

两边关于x求一阶导y'*e^(x+y)-y'sinx-ycosx=0y'=ycosx/(e^(x+y)-sinx)

再答：是（x+y）^2还是x+y^2再问：是前者再问：第一道题你算错了吧。再答：为啥。。。。再问：再问：这个是答案。再答：第二个你把分子分母倒一下。。。。我看看。。？再问：？？再问：再问：第二道题再答

[d(ylny)/dy]*dy/dx-1+dy/dx=0dy/dx=1/(2+lny)

设y=y(x)由方程ysinx=cos(x-y)所确定,则y'(0)=x=0时cos(-y)=cosy=0,故y=π/2+2kπ,k∈ZF(x,y)=ysinx-cos(x-y)=0dy/dx=-(&

dy/ylny=dx/x两边积分得lnlny=lnx+C1lny=C2e^x再问：后面那题呢？再答：y=x(-1≤x≤1)再问：cosxsinydy=cosysinxdx,Y|(x=0)=45°求初始

参考答案:停车坐爱枫林晚,霜叶红于二月花.

两边对x求导：dy/dxsinx+ycosx-sin(x-y)(1-dy/dx)=0,将x=π/2带入已知方程得到y,再把x、y带入上式求得结果再问：x=π/2带入已知方程得到y。。。我算不出这个y

ysinx-cos(x+y)=0,两边对x求导,得y'sinx+ycosx+(1+y')sin(x+y)=0,解得y'=-[ycosx+sin(x+y)]/[sinx+sin(x+y)]dy/dx=y

∵ylnydx+(x-lny)dy=0∴ylnydx/dy+x=lny.(1)∴原方程与方程(1)同解用常数变易法求解方程(1)∵ylnydx/dy+x=0==>dx/x=-dy/(ylny)==>d

这很简单啊y'sinx=ylnydy/(ylny)=sinxdxd(lny)/lny=sinxdx两边积分得到ln(lny)=-cosx+C,C是任意常数

应用复合函数求导方法,y′sinx+ycosx+（1+y′）sin（x+y）=0,（sinx+sin（x+y））y′+ycosx+sin（x+y）=0,y′=-（ycosx+sin（x+y））/（si

两边对x求导y'*sinx+ycosx-[-sin(x+y)*(1+y')]=0y'(sinx+sin(x+y))=y(1-cosx)y'=[1-cosx]/[sinx+sin(x+y)]0/0所以需

可分离变量型,原微分方程可化为dx/(1+x^2)=dy/(ylny),两边同时积分J1/(1+x^2)dx=J1/(lny)d(lny),得lnlny=arctanx+C1得通解lny=Ce^(ar

两边同时对y积分得d(yy')=d(0.5y^2(lny-0.5))y'=0.5ylny-1/4y+c1/y积分得y=1/4y^2lny-1/4y^2+C1lny+C2

么么　嘿嘿　爱死你啦

数列1/1*2+1/2*3+…+1/n(n+1)的sn=1-1/2+1/2-1/3+----+1/n-1/(n+1)=1-1/(n+1)1-1/(n+1)中的1-是怎么得出的?1/n-的n取1吗,你不

两边求导：y'sinx+ycosx+sin(x+y)*(1+y')=0令x=0,y=π/2：π/2+1+y'=0y'=-(π/2+1)dy=-(π/2+1)dx