求y∧2dx (x 1)dy=0通解,并满足初值条件y(0)=1的特解-搜答案-搜搜考试网

∵(x+y)dy+(x-y)dx=0==>(1+y/x)dy+(1-y/x)dx=0设y=xt,则dy=tdx+xdt∴(x+y)dy+(x-y)dx=0==>(1+t)(tdx+xdt)+(1-t)

x^2y''-2xy'=y'^2-(2xy'-x^2y'')/y'^2=1(x^2/y')'=-1两边积分：x^2/y'=-x+C1y'=x^2/(-x+C1)=(x^2-C1x+C1x-C1^2+C

y=xe^y 求dy/dx

(1+y)dx-(1-x)dy=0(1+y)dx=(1-x)dy[1/(1-x)]dx=[1/(1+y)]dyd(ln(1-x))=d(ln(1+y))ln(1-x)+C1=ln(1+y)（C1为任意

y和x都是t的函数,而y又是x的函数.dy/dt求导直接=y''(t)dy/dx就不能直接出答案了,先转化为(dy/dt)*(dt/dx)然后求导,这样求出来都是关于t的导数

y=x∧x,求dy/dx,

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y=x∧sinx,求dy/dx,

图片中有详细的过程,楼主看完满意的话别忘了选为满意答案

(x^2)dy+（y^2)dx=dx-dy

（x^2+1)dy=(1-y^2)dxdy/(1-y)(1+y)=dx/(x^2+1)1/2lnl(y-1)/(y+1)l=arctanx+c再问：在帮我一个，我给再加五分，y′=y，y（0）=1.谢

∵dx+(x+y^2)dy=0==>e^ydx+xe^ydy+y^2e^ydy=0(等式两端同乘e^y)==>e^ydx+xd(e^y)+y^2e^ydy=0==>d(xe^y)+d((y^2-2y+

y=(lnx)∧y,求dy/dx=?

隐函数求导的结果里面通常含x和y.lny=y*ln(lnx)求导：y'/y=y'*ln(lnx)+y*1/[x*lnx]=>y'=y^2/[x*lnx*(1-y*ln(lnx))]再问：可是这题的答案

再答：另一个自己做再问：谢谢，明白思路了

dy/dx=-cosx;dy/dx|x=0=cos0=1.

e∧x+y=xy,求dy/dx

y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即：dy/dx=y/[2πycos(πy²

令p=dy/dx,则d^2y/dx^2=pdp/dy代入方程：pdp/dy-e^yp=0dp/dy=e^ydp=e^ydy积分：p=e^y+cdy/dx=e^y+cdy/(e^y+c)=dxd(e^y

求dy/dx=(x/y)+cos²(x/y)通解令x/y=u,则y=x/u,dy/dx=[u-x(du/dx)]/u²,代入原式得：[u-x(du/dx)]/u²=u+c

dy/dx=(x-y+5)/(x+y-2)=[(x+3/2)-(y-7/2)]/[(x+3/2)+(y-7/2)]令v=y-7/2,u=x+3/2,原方程化为dv/du=(u-v)/(u+v)变为齐次

y=xe^y,求dy/dx |x=0

y'=(xe^y)'=x'e^y+x(e^y)'=e^y+xe^yy'y‘=e^y/(1-e^y)∴dy/dx=e^y/(1-e^y)x=0好象没有一个确定的值