Question

求解微分方程(1-x)y''+xy'-y = -(x-1)^2

Answer 1

∵(1-x²)y'+xy=0 ==>dy/y=-xdx/(1-x²)
==>dy/y=(1/2)d(1-x²)/(1-x²)
==>ln│y│=(1/2)ln│1-x²│+ln│C│ (C是积分常数)
==>y=C√(1-x²)
∴齐次方程(1-x²)y'+xy=0的通解是y=C√(1-x²) (C是积分常数)
于是,设微分方程(1-x²)y'+xy=1的解为 y=C(x)√(1-x²) (C(x)是关于x的函数)
∵y'=C'(x)√(1-x²)-C(x)x/√(1-x²)
代入原方程得C'(x)=1/√(1-x²)³
∴C(x)=∫dx/√(1-x²)³
=∫costdt/cos³t (设x=sint,则tant=x/√(1-x²)
=-∫dt/cos²t
=-∫sec³tdt
=tant+C (C是积分常数)
=x/√(1-x²)+C
∴y=[x/√(1-x²)+C]√(1-x²)
=x+C√(1-x²)
∵y(0)=1 ==>C=1
故原微分方程的解是 y=x+√(1-x²)
heanmen | 2010-06-02