求微分方程xy''=(1+2x^2)y'的通解是,

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xy''=y'+2x^2y'
xy''-y'=2x^2y' 两边同除以x^2
(xy''-y')/x^2=2y'
(y'/x)=2y+c
y'/(2y+c)=x
1/2 ln(2y+c1)=1/2 x^2+c2
ln(2y+C1)=x^2+C2
2y+C1=e^(x^2+C2) =C2* e^(x^2)
y= C1 * e^(x^2)+C2
======以下答案可供参考======
供参考答案1:
xy''=(1+2x^2)y'
y''=(1/x+2x)y'
解得y'=C1e^(lnx+x^2)=C1xe^(x^2)
y=C1∫xe^(x^2)dx
=(C1/2)e^(x^2)+C2
=Ce^(x^2)+C2