Question

y的二阶导-y的一阶导-y的²＝0的通解是什么

Answer 1

y''
+
y'
=
e^x.......................(1)
y''
+
y'
=
0............................(2)
对应的特征根：
s1＝0.........s2＝－1
(1)的特解：
y1＝0.5e^x...........................(3)
(2)的通解：
y*
＝
c1
+
c2
e^(-x).........(4)
(1)的通解:
y
=
y*
+
y1
即：
y
=
c1
+
c2
e^(-x)
+
0.5
e^(x).............(5)
c1,
c2
为积分常数，由初始条件确定。

Answer 2

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