y-xy'=b(1-x^2y'),y|x=1 =1的特解
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y-xy''=b(1-x^2y''),y|x=1=1的特解y-xy''=b(1-x^2y''),y|x=1=1的特解y-xy''=b(1-x^2y''),y|x=1=1的特解y-xy''=b(1-x²y''
y-xy'=b(1-x^2y'),y|x=1 =1的特解
y-xy'=b(1-x^2y'),y|x=1 =1的特解
y-xy'=b(1-x^2y'),y|x=1 =1的特解
y-xy'=b(1-x²y')
化成y'+p(x)y=q(x) 的形式,即有
p(x)=1/(bx²-x),q(x)=b/(bx²-x)
p(x)积分=ln[(bx-1)/x]
q(x)e^ln[(bx-1)/x]=b/x²积分=-b/x
所以通解就是
y=e^{-ln[(bx-1)/x]}*(-b/x + C)
=x/(bx-1)*(-b/x + C)
=-b/(bx-1)+Cx/(bx-1)
把x=1,y=1代入上式得
C=2b-1
所以特解就是
y=-b/(bx-1)+x(2b-1)/(bx-1)
也可表示为 y(bx-1)=-b+x(2b-1)
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