Question

微分方程xy"＋y＝0怎么求？

Answer 1

解法一：∵xy''+y'=0 ==>xdy'/dx=-y'
==>dy'/y'=-dx/x
==>ln│y'│=-ln│x│+ln│C1│ (C1是积分常数)
==>y'=C1/x
==>y=C1ln│x│+C2 (C2是积分常数)
∴原方程的通解是y=C1ln│x│+C2 (C1,C2是积分常数)；
解法二：∵令t=ln│x│,则xy'=dy/dt,x²y''=d²y/dt²-dy/dt
代入原方程得 d²y/dt²-dy/dt+dy/dt=0
==> d²y/dt²=0
==>dy/dt=C1 (C1是积分常数)
==>y=C1t+C2 (C2是积分常数)
==>y=C1ln│x│+C2
∴原方程的通解是y=C1ln│x│+C2 (C1,C2是积分常数).

追问

能看清楚题目么！是显含y与x的