ysinx=ylny在y(π 2)=e
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(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)
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)
如果有用请及时采纳,
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
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
反函数是表达不出来的,只能用隐函数求导法.即求该点的两阶导数.
ylny应该看成复合函数一阶:y'lny+y*1/y*y'-1+y'=0所以y'=1/(2+lny)=y/(y+x)二阶:y''=-1/(2+lny)^2*(2+lny)'=-y'/[y(2+lny)
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
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