已知(1-x^2)f"(x)+xf'(x)=x/(x+1)已知1.(1-x^2)f"(x)+xf'(x)=x/(x+1)在（-1,1）上成立2.f(x)在（-1,1）上二阶可导3.f(0)=04.f(0.5)=1.5+0.5*(pi/6)^2+pi/6求f(x)

已知(1-x^2)f"(x)+xf'(x)=x/(x+1)已知1.(1-x^2)f"(x)+xf'(x)=x/(x+1)在（-1,1）上成立2.f(x)在（-1,1）上二阶可导3.f(0)=04.f(0.5)=1.5+0.5*(pi/6)^2+pi/6求f(x)
已知(1-x^2)f"(x)+xf'(x)=x/(x+1)
已知
1.(1-x^2)f"(x)+xf'(x)=x/(x+1)
在（-1,1）上成立
2.f(x)在（-1,1）上二阶可导
3.f(0)=0
4.f(0.5)=1.5+0.5*(pi/6)^2+pi/6
求f(x)

已知(1-x^2)f"(x)+xf'(x)=x/(x+1)已知1.(1-x^2)f"(x)+xf'(x)=x/(x+1)在（-1,1）上成立2.f(x)在（-1,1）上二阶可导3.f(0)=04.f(0.5)=1.5+0.5*(pi/6)^2+pi/6求f(x)
用分部积分配方法：.
(1-x^2)f "(x) + xf '(x) ＝ x/(x+1) 方程两边乘以(1-x^2)^(-3/2)
==> (1-x^2)^(-1/2) f "(x) + x(1-x^2)^(-3/2) f '(x) ＝ x(1-x^2)^(-3/2)/(x+1)
==> [ (1-x^2)^(-1/2) f '(x) ] ' ＝ x(1-x^2)^(-3/2)/(x+1)
==> (1-x^2)^(-1/2) f '(x) ＝ ∫ x(1-x^2)^(-3/2)/(x+1) dx
==> (1-x^2)^(-1/2) f '(x) ＝ ∫ (x-x^2)(1-x^2)^(-5/2) dx
==> (1-x^2)^(-1/2) f '(x) ＝ (1-x^3)(1-x^2)^(-3/2) / 3 + C1
==> f '(x) ＝ (1-x^3)(1-x^2)^(-1) / 3 + C1*(1-x^2)^(1/2)
==> f (x) ＝ ∫ [ (1-x^3)(1-x^2)^(-1) / 3 + C1*(1-x^2)^(1/2) ] dx
==> f (x) ＝ ∫ [ (1+x+x^2)/(3(1+x)) + C1*(1-x^2)^(1/2) ] dx
==> f (x) ＝ x^2/6 + ln(1+x) /3 + C1*[ x(1-x^2)^(1/2)/2 + arcsin(x)/2 ] + C2
==> f (x) ＝ x^2/6 + ln(1+x) /3 + C1*[ x(1-x^2)^(1/2) + arcsin(x) ] + C2 （C1为常数,1/2可去掉)
==> f (x) ＝ x^2 / 6 + ln(1+x) /3 + C1*[ x√(1-x^2) + arcsin(x) ] + C2 （到此步已验证正确)
根据初始条件解得:C2＝0,C1＝?
f(0.5)给的有点问题,C1的结果很复杂,不再计算.

设p(x)=f'(x) 则f''(x)=p'(x)
(1-x^2)p'(x)+xp(x)=x/(x+1)
(1-x^2)p'(x)=-xp(x)
p'/p=-x/(1-x^2)
lnp=1/2ln|1-x^2|+lnC
p=C(1-x^2)^(1/2)
p'=C'(1-x^2)^(1/2)-2xC/2(1-x^2)^(-1/2)
(1-x...

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设p(x)=f'(x) 则f''(x)=p'(x)
(1-x^2)p'(x)+xp(x)=x/(x+1)
(1-x^2)p'(x)=-xp(x)
p'/p=-x/(1-x^2)
lnp=1/2ln|1-x^2|+lnC
p=C(1-x^2)^(1/2)
p'=C'(1-x^2)^(1/2)-2xC/2(1-x^2)^(-1/2)
(1-x^2)^(3/2)C'-xC(1-x^2)^(1/2)+xC(1-x^2)^(1/2)=x/(x+1)
C'=x(1-x^2)^(-3/2)/(x+1)=x(1-x)^(-3/2)(1+x)^(-5/2)
C=-1/3*(1-x)^(1/2)*(x^2+x+1)/(1+x)^(3/2)/(x-1)+C
f'(x)=-1/3*(1-x)^(1/2)*(x^2+x+1)/(1+x)^(3/2)/(x-1)*(1-x^2)^(1/2)+C(1-x^2)^(1/2)
=(x^2+x+1)/(3+3*x)+C(1-x^2)^(1/2)
f(x)=1/6*x^2+1/3*ln(3+3*x)+1/2*C*x*(1-x^2)^(1/2)+1/2*C*arcsin(x)+C1
∵f(0)=1/3*ln(3)+C1=0
∴C1=-ln3/3
∵f(0.5)=1/6*1/4+1/3ln(9/2)+C*1/4*√3/2+C*1/2*pi/6-ln3/3
=1/24+2/3ln3-1/3ln2-ln3/3+( √3/8+pi/12)C
=1.5+0.5*(pi/6)^2+pi/6
∴ C=[1.5+0.5*(pi/6)^2+pi/6-1/24-ln3/3+ln2/3]/(√3/8+pi/12)
f(x)=1/6*x^2+1/3*ln(3+3*x)+1/2*C*x*(1-x^2)^(1/2)+1/2*C*arcsin(x)+C1
C1=-ln3/3 C=[1.5+0.5*(pi/6)^2+pi/6-1/24-ln3/3+ln2/3]/(√3/8+pi/12)
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