问题描述： y"=3根号y,x=0,y=1 x=0,y'=2 求特解 1个回答 分类：数学 2018-03-20 问题解答： 我来补答 ∵y"=3√y ==>y'dy'/dy=3√y ==>y'dy'=3√ydy ∴y'^2/2=2y^(3/2)+C1 (C1是积分常数) ∵当x=0时,y=1,y'=2 ==>C1=0 ∴y'^2/2=2y^(3/2) ==>y'^2=4y^(3/2) ==>y'=±2y^(3/4) ==>dy/dx=±2y^(3/4) ==>dy/y^(3/4)=±2dx ==>4y^(1/4)=C2±2x (C2是积分常数) ∵当x=0时,y=1 ==>C2=4 ∴4y^(1/4)=4±2x ==>y=(1±x/2)^4 故原方程满足所给初始条件的特解是y=(1±x/2)^4. 展开全文阅读
