y=2x^2+cosx,(d^2y)/(dx^2)=?
已知y=cosx^2,求d^y/dx^2;d^y/dx^3..
求dy/dx=x/y+(cosx/y)^2通解
已知y=x^cosx/2,求dy/dx,
dy/dx=y^2cosx
(cosx+1/y)dx+(1/y-x/y^2)dy=0
求解微分方程dy/dx +y=y^2(cosx-sinx)
d/dx((x^2+y^2)^2)=d/dx(x^2-y^2),求(化简)dy/dx=?
d^2y /dx^2 - 24 =0 dy/dx -2y = e^-x
y=x^2*4^x+lnx/cosx,求dy/dx
y"-2y'+2y=x*(e^x)*cosx
试从dx/dy=1/y'导出:d^2x/dy^2=-y''/(y')^3 题目中关于d[1/y']/dx}*[dx/dy
从(dx)/(dy)=1/y '导出:(d^2x)/(dy^2)=-y''/(y')^3