Question

微分方程xy''=y'-x(y')^2的通解为 y''+2y'=0求解，给出步骤

Answer 1

y'-2y=0,
特征方程为r-2=0,得r=2,通解为y=ce^(2x)
y'-2y=x,
r=2
,设特解为y*=ax+b,
则a-2ax-2b=x,得：a-2b=0,
-2a=1,得a=-1/2,
b=-1/4
故通解为y=ce^(2x)-(1/2)x-1/4
y"+y=0，特征方程r²+1=0,得r=i,
-i,
通解为y=c1cosx+c2sinx
y"+y=x,
设特解y*=ax+b,
代入方程得：ax+b=x,
得a=1,
b=0,
故通解为y=c1cosx+c2sinx+x

Answer 2

1
xy''=y'-x(y')^2
xy''/(y')^2-y'/(y')^2=-x
d(x/y')/dx=(xy''-y')/(y')^2
d(x/y')=-xdx
x/y'=-x^2/2
1/y'=-x/2
dx/dy=-x/2
dx/x=-dy/2
lnx=-y/2+C
y=-ln(x^2)+C'

2
y''+2y'=0
y''/y'=-2
(lny')'=2
d(lny')/dx=2
lny'=2x+C
y'=Ce^(2x)
dy/dx=Ce^(2x)
dy=Ce^(2x)dx
y=C'e^(2x)