y的二阶导-y的一阶导-y的눀＝0的通解是什么

解：
令：G(x,y)=x²-xy+y²-1=0
根据隐函数定义：
对上式求关于x的偏导：
G'x=2x-y
对上式求关于y的偏导：
G'y=2y-x
∴
dy/dx
= - G'x/G'y
= (2x-y)/(x-2y)
对z=x²+y²求关于x的偏导
dz/dx
=2x+2y·(dy/dx)
=2x+2y·(2x-y)/(x-2y)
=(2x²-4xy+4xy-2y²)/(x-2y)
=2(x²-y²)/(x-2y)
d²z/dx²
={2[2x-2y·(dy/dx)]·(x-2y)-2(x²-y²)·[1-2(dy/dx)]}/(x-2y)²
={[4x(x-2y)-4y·2(x²-y²)]-2(x²-y²)·[1-2·2(x²-y²)/(x-2y)]}/(x-2y)²
={[4x(x-2y)²-4y·2(x²-y²)(x-2y)]-2(x²-y²)·[1-2·2(x²-y²)]}/(x-2y)³
=(4x³+16xy²-16xy-8yx³+16x²y²+8xy³+16y·y³-2x²-2y²+8x·x³+8yy³-16x²y²)/(x-2y)³
=(4x³+16xy²-16xy-8yx³+8xy³+24y^4-2x²-2y²+8x^4)/(x-2y)³

特征根方程r²-r-1=0
r=(1±√5)/2
y=C1e^((1+√5)x/2)+C2e^((1-√5)x/2)