Question

xy''-y'=1的通解？

网上计算器说的是C1+C2*x^2+x可是我求得yb=-1/4x而不是x求步骤解

Answer 1

方法如下，
请作参考：

Answer 2

简单计算一下即可，答案如图所示

Answer 3

xy'' - y' = 1, 记 y' = p, 则 xdp/dx = 1+p, dp/(1+p) = dx/x,
ln(1+p) = lnx+ln(2C1), 1+p = 2C1x, y' = 2C1x-1
y = C1x^2 - x + C2

Answer 4

p(x) = -1/x
∫p(x) dx = -ln|x| +C
e^[∫p(x) dx] =1/x
xy''-y'=1

y''- (1/x)y'= 1/x
两边乘以 1/x
(1/x)[y''- (1/x)y']= 1/x^2
d/dx (y'/x)= 1/x^2
y'/x= ∫(1/x^2) dx
=-1/x + C1'
y' =-1 + C1'.x
y=∫[-1 + C1'.x] dx
= -x +C1'x^2/2 +C2
=-x + C1.x^2 +C2