求下列微分方程的通解：y"=1/(y)^0.5

Question

求下列微分方程的通解：y"=1/(y)^0.5

Answer 1

y'=p
y''=pdp/dy
pdp/dy=y^(-0.5)
pdp=y^(-0.5)dy
两边积分：
1/2p^2=1/(1-0.5) * y^(1-0.5)+c
1/2p^2=2y^(0.5)+c
1/4p^2=y^(0.5)+C
1/2p=土(y^(0.5)+C)^(0.5)
y'=土2(y^(0.5)+C)^0.5
由于
dy/根(根y+C)中,设根(根y+C)=t t^2=根y+C (t^2-C)^2=y
2(t^2-C)*2tdt=dy
故：dy/根(根y+C)＝1/t * 4*(t^3-Ct)dt=(4t^2-4C)dt
故：(4t^2-4C)dt=土2dx
4/3t^3-4Ct=土2x+C3
2/3t^3-2Ct=土x+C2
2/3(y^0.5+C)^1.5 - 2C(y^0.5+C)^0.5=土x+C2
要命,自已算一下!