求微分方程xy''+y'=0的通解（提示：可降階）

數(shù)學(xué)人氣：266 ℃時(shí)間：2020-06-19 04:59:19

優(yōu)質(zhì)解答

解法一：∵xy''+y'=0 ==>xdy'/dx=-y'
==>dy'/y'=-dx/x
==>ln│y'│=-ln│x│+ln│C1│ (C1是積分常數(shù))
==>y'=C1/x
==>y=C1ln│x│+C2 (C2是積分常數(shù))
∴原方程的通解是y=C1ln│x│+C2 (C1,C2是積分常數(shù))；
解法二：∵令t=ln│x│,則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是積分常數(shù))
==>y=C1t+C2 (C2是積分常數(shù))
==>y=C1ln│x│+C2
∴原方程的通解是y=C1ln│x│+C2 (C1,C2是積分常數(shù)).