第3个回答 2021-09-02
p(x) =-[2x/(1+x^2)]
∫p(x) dx = ∫-[2x/(1+x^2)] dx = -ln|1+x^2|
e^[∫p(x) dx] = 1/(1+x^2)
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(1+x^2)y''=2xy'
(1+x^2)y''-2xy' =0
y''-[2x/(1+x^2)]y' =0
两边乘以 1/(1+x^2)
[ 1/(1+x^2) ].[y''-[2x/(1+x^2)]y' ]=0
d/dx [y'/(1+x^2)] =0
y'/(1+x^2) = C1
y' =C1.(1+x^2)
y=∫ C1.(1+x^2) dx
= C1.x + (1/3)C1.x^3 +C2
y(0) =1 => C2=1
y' = C1 +C1.x^2
y'(0) =3 => C1=3
ie
y= 3x + x^3 +1