高数常微分方程解初值问题! 求详细过程
1个回答
展开全部
let
u= xy
du/dx = xdy/dx + y
/
dy/dx + y/x = 2(lnx).y^2
xdy/dx + y = 2x(lnx).y^2
du/dx =2x^3.(lnx).u^2
∫du/u^2 = ∫ 2x^3 .(lnx) dx
-1/u = (1/2)∫ (lnx) dx^4
= (1/2)x^4.lnx -(1/2)∫ x^3 dx
=(1/2)x^4.lnx -(1/8)x^4 + C
y(1) = 1
-1 = -1/8 + C
C = -7/8
-1/u =(1/2)x^4.lnx -(1/8)x^4 -7/8
-1/(xy) = (1/2)x^4.lnx -(1/8)x^4 -7/8
y = -1/[(1/2)x^5.lnx -(1/8)x^5 -(7/8)x ]
=-8/[4x^5.lnx -x^5 -7x ]
u= xy
du/dx = xdy/dx + y
/
dy/dx + y/x = 2(lnx).y^2
xdy/dx + y = 2x(lnx).y^2
du/dx =2x^3.(lnx).u^2
∫du/u^2 = ∫ 2x^3 .(lnx) dx
-1/u = (1/2)∫ (lnx) dx^4
= (1/2)x^4.lnx -(1/2)∫ x^3 dx
=(1/2)x^4.lnx -(1/8)x^4 + C
y(1) = 1
-1 = -1/8 + C
C = -7/8
-1/u =(1/2)x^4.lnx -(1/8)x^4 -7/8
-1/(xy) = (1/2)x^4.lnx -(1/8)x^4 -7/8
y = -1/[(1/2)x^5.lnx -(1/8)x^5 -(7/8)x ]
=-8/[4x^5.lnx -x^5 -7x ]
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询