y^2dx-(2xy+3)dy=0求通解

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/06 00:30:24
y^2dx-(2xy+3)dy=0求通解
xN@_evkccKah(EU5REcpJUB/i. ZpMb73w=, 7g2wG6o',"6 ⓎQ^ fMnW,VtVok;##^O4Qe6ُ$}[uwYi Ѐ偬}u0 j/ʉE"t9+ AG[`I]&,L0 =e5ejcn1MdHO |[-muq*GelQ݄p.tMW@]i쐧i0-'EOU _W B1PE`d4NW2aT"+p-R9L J|0k W8p_ZxY}13q

y^2dx-(2xy+3)dy=0求通解
y^2dx-(2xy+3)dy=0求通解

y^2dx-(2xy+3)dy=0求通解
y^2 dx - (2xy+3) dy = 0
dy / dx = y^2 / (2xy+3)
当y=0时,微分方程显然成立,所以y=0是原方程的一个解
当y不=0时,
-y^(-2)dy / dx = -1 / (2xy+3)
d(1/y) / dx = -1/y / (2x+3/y)
令 1/y = t
则 dt / dx = -t / (2x+3t)
dt / dx = -1 / (2x/t + 3)
dx / dt = -(2x/t + 3)
令 s = x/t 则 x = ts dx / dt = s + t×ds/dt
s + t×ds/dt = -2s - 3
t×ds/dt = -3s - 3
1/(s+1)×ds = -3/t×dt
上式为变量分离方程,两边积分
ln(s+1) = -3lnt + C‘
s+1 = e^C’ × t^(-3)
x/t + 1 = e^C‘ × t^(-3)
xy + 1 = e^C’ × y^3
e^C‘ × y^3 - xy -1 = 0 (C’为任意常数)
Cy^3 - xy -1 = 0 (C为任意正数)
所以,原微分方程的通解为 y=0及Cy^3 - xy -1 = 0 (C为任意正数)

dx/dy-2/y*x=3/y^2
x=e^(-∫-2/y*dy)[∫3/y^2*e^(∫-2/y*dy)dy+C]
=y^2[∫3/y^2*(1/y^2)dy+C]
=y^2[∫3/y^4*dy+C]
=y^2(-1/y^3+C)