求微分方程y''+(2x/x^2+1)y'-2x=0
1个回答
y''+2xy'/(x^2+1)-2x=0
(x^2+1)y''+2xy'-2x(x^2+1)=0
x=tanu y'=dy/dx=cosu^2 dy/du u=arctanx du/dx=1/(1+x^2)
y''=dy'/dx=d[cosu^2 dy/du]/du*du/dx
=[-2sin2udy/du+cosu^2 d^2y/du^2 ] *(1/(1+x^2))
(1+x^2)y''=-2sin2udy/du+cosu^2d^2y/du^2
2xy'=sin2udy/du
2x(x^2+1)=2tanu*secu^2=2sinu/cosu^3
cosu^2d^2y/du^2-2sinu/cosu^3=0
d^2y/du^2=2sinu/cosu^5
dy/du=∫2sinudu/cosu^5
=(1/2)(1/cosu^4)+C1
y=∫[(1/2)(1/cosu^4)+C1]du
=C1u+(1/4)secu^2tanu+(1/4)tanu
=C1arctanx+(1/4)x*(1+x^2)+(1/4)x
∫secx^4dx=∫secx^2dtanx=secx^2 tanx-∫tanx^2secx^2dx
=secx^2tanx-∫secx^4dx+∫secx^2dx
2∫secx^4dx=secx^2tanx+tanx
∫secx^4dx=(1/2)secx^2tanx+(1/2)tanx
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