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求大佬帮忙解一下这道微分方程题
5个回答
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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)
//
(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
∫p(x) dx = ∫-[2x/(1+x^2)] dx = -ln|1+x^2|
e^[∫p(x) dx] = 1/(1+x^2)
//
(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
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解:∵微分方程为(1+x²)y"=2xy',化为
y"/y'=2x/(1+x²) ∴有ln|y'|=
ln|1+x²|+ln|a|(a为任意非零常数)
y'=a(1+x²),y=ax+ax³/3+b(b为任意常数) ∵y(0)=1,y'(0)=3 ∴有a=3,b=1
∴微分方程的解y=3x+x³+1
y"/y'=2x/(1+x²) ∴有ln|y'|=
ln|1+x²|+ln|a|(a为任意非零常数)
y'=a(1+x²),y=ax+ax³/3+b(b为任意常数) ∵y(0)=1,y'(0)=3 ∴有a=3,b=1
∴微分方程的解y=3x+x³+1
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