y"=3根号y,x=0,y=1x=0,y'=2求特解

Question

y"=3根号y,x=0,y=1 x=0,y'=2 求特解

Answer 1

∵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.