设 p=y'=dy/dx,
则 y²y/dx²=dp/dx=(dp/dy) * (dy/dx)
=pdp/dy,
方程化为 2pdp=2dy/y,
积分得 p²=ln(Cy²),
所以 y'=±√ln(Cy²),
分离变量得 dy/±√ln(Cy²) = dx,
再积分即可(感觉思路不对啊。。)
解:∵微分方程y''=1/y,化为2y'y''=2y'/y
∴有y'²=lny²+lna²(a为任意非零常数)
y'=±√[ln(a²y²)],
dy/√[ln(a²y²)]=±dx,方程的通解
x=∫dy/[ln(a²y²)]+c(c为任意常数)