微分方程y#39;#39;-3/2y^2=0满足初始y(0)=1,y#39;(0)=1

问题：微分方程y#39;#39;-3/2y^2=0满足初始y(0)=1,y#39;(0)=1

答案：↓↓↓

范金华的回答：

网友采纳　　设y'=p(y),则y''=dp/dx=dp/dy*p,原方程化为2pdp=3y^2dy,∴p^2=y^3+c,∴y'=土√(y^3+c),把初始条件代入得c=0,取正号,dy/√（y^3)=dx,-2/√y=x+c1,把初始条件代入得c1=-2,∴-2/√y=x-2,√y=2/(2-x),∴y=[2/(2-x)}^2,(x...