y″+a^2/y^2=0的通解为多少啊 y″=1/根号ay 的通解

1y'=dy/dx=py''=dp/dx=dp/dy*dy/dx=pdp/dypdp/dy+a^2/y^2=0pdp=-a^2dy/y^2p^2=a^2/y+Cp=√(a^2/y+C) 或 p=-√(a^2/y+C)dy/dx=√(a^2/y+C)ydy/√(a^2+Cy^2)=dxdy^2/√(1+Cy^2/a^2)=adx通解√(1+Cy^2/a^2)=(a^3/C)x+C1...p^2=a^2/y+Cp=√(a^2/y+C)或 p=-√(a^2/y+C)dy/dx=√(a^2/y+C)ydy/√(a^2+Cy^2)=dxdy^2/√(1+Cy^2/a^2)=adx通解√(1+Cy^2/a^2)=(a^3/C)x+C1 或 √（1+Cy^2/a^2)= (-a^3/C)x+C1 这几步不太清楚能再算一遍吗非常感谢啊谢谢，确有误算，修正如下：p^2=a^2/y+Cp=√(a^2/y+C)或 p=-√(a^2/y+C)dy/dx=√(a^2/y+C)=√(a^2+Cy)/√y√ydy/√(1+Cy/a^2)=adxax=∫√ydy/√(1+Cy/a^2)=(a/√C)∫√(Cy/a^2)dy/√(1+Cy/a^2)=(a^3/√C^3) ∫√(Cy/a^2) d(Cy/a^2)/√(1+Cy/a^2)=(a^3/√C^3)* [√(1+Cy/a^2)√(Cy/a^2) -ln|√(1+Cy/a^2) +√(Cy/a^2)] +CCy/a^2=u∫√udu/√(1+u)u=tanv^2du=2tanvsecv^2dv=∫2tanv^2secv^2dv/secv=∫2tanv^2secvdv=∫2(secv^2-1)secvdv=∫2secvdtanv-2∫secvdv =secvtanv-ln|secv+tanv|∫secvdtanv=secvtanv-∫tanv^2secvdv =secvtanv-∫secvdtanv+∫secvdv∫secvdtanv=(1/2)secvtanv+(1/2)∫secvdv∫secvdv=ln|secv+tanv|