我有个疑问啊,就是那道设x=e^(-t) 试变换方程x^2 d^2y/dx^2 +xdy/dx+y=0,把X
作变换u=tany,x=e的t次幂 试将方程 x^2d^2y/dx^2+2x^2(tany)(dy/dx)^2+xdy/
求 [y+x^2*e^(-x)]dx-xdy=0 的通解
xdy/dx=y+x^2 求通解
求微分方程xdy-(2y+x^4)dx=0.,
设函数y=y(x)由x=1-e^t和y=t+e^-t确定,求dy/dx和d^2y/dx^2
求解微分方程 [y-x(x^2+y^2)]dx-xdy=0
令x=cost,变换方程d^2y/dx^2-x/(1-x^2)*dy/dx+y/(1-x^2)=0
证明x^2(d^2y/dx^2)+a_1x(dy/dx)+a_2y=0 ,令x=e^t,方程可化成d^2y/dt^2+(
方程ydx-xdy=(x^2+y^2)dx的通解
已知 x=e^t ,dy/dx=dy/xdt .分析变换具体步骤 d^2y/dx^2=(d^2y/dt^2-dy/dt)
求微分方程的通解[y+(x^2+y^2)^1/2]dx-xdy=0
设y=y(x)是由方程xy+e^y=y+1所确定的隐函数,求d^2y/dx^2 x=0