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切线方程
y=kx+b
上式中k就是y’
b 就是x的平方加y的平方再开平方。
x=1/2时y=0是初始条件。
y=kx+b
上式中k就是y’
b 就是x的平方加y的平方再开平方。
x=1/2时y=0是初始条件。
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(6)
u=e^x
y'= dy/dx = (dy/du).(du/dx) = u.dy/du
y''
= d/dx (y')
=d/du (y') . (du/dx)
=u.d/du (y')
=u.d/du ( u.dy/du )
=u[ dy/du + u. d^2y/du^2 ]
=u.dy/du + u^2. d^2y/du^2
y'' -(2e^x +1)y' + e^(2x).y = e^(3x)
y''-(2u+1)y' + u^2.y = u^3
[u.dy/du + u^2. d^2y/du^2] -(2u+1).[u.dy/du] +u^2.y = u^3
u^2. d^2y/du^2 -2u^2 . dy/du +u^2.y = u^3
d^2y/du^2 -2. dy/du +y = u
The aux. equation
p^2-2p+1=0
p=1
let
yg=(Au+B)e^u
yp = Cu +D
yp'=C
yp''= 0
yp''-2yp+yp=u
-2C +Cu+D =u
Cu +(-2C+D)=u
C=1 and -2C+D=0
C=1 and D=2
yp = Cu +D = u +2
通解
y
=yg+yp
=(Au+B)e^u + u+2
=(Ae^x+B)e^(e^x) + e^x+2
u=e^x
y'= dy/dx = (dy/du).(du/dx) = u.dy/du
y''
= d/dx (y')
=d/du (y') . (du/dx)
=u.d/du (y')
=u.d/du ( u.dy/du )
=u[ dy/du + u. d^2y/du^2 ]
=u.dy/du + u^2. d^2y/du^2
y'' -(2e^x +1)y' + e^(2x).y = e^(3x)
y''-(2u+1)y' + u^2.y = u^3
[u.dy/du + u^2. d^2y/du^2] -(2u+1).[u.dy/du] +u^2.y = u^3
u^2. d^2y/du^2 -2u^2 . dy/du +u^2.y = u^3
d^2y/du^2 -2. dy/du +y = u
The aux. equation
p^2-2p+1=0
p=1
let
yg=(Au+B)e^u
yp = Cu +D
yp'=C
yp''= 0
yp''-2yp+yp=u
-2C +Cu+D =u
Cu +(-2C+D)=u
C=1 and -2C+D=0
C=1 and D=2
yp = Cu +D = u +2
通解
y
=yg+yp
=(Au+B)e^u + u+2
=(Ae^x+B)e^(e^x) + e^x+2
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