Question

高数题一道求微分方程x^2y'+xy=y^2满足初始条件y（1）=1的特解.

Answer 1

x^2y'+xy = y^2 ， 是齐次方程， 令 y = xu， 则 y' = u+xu'，
原微分方程化为 2ux^2 + x^3u' = x^2u^2,
x ≠ 0 时 2u + xu' = u^2， xdu/dx = u(u-2)
du/[u(u-2)] = dx/x, [1/(u-2) - 1/u]du = 2dx/x,
ln[(u-2)/u] = 2lnx + lnC
(u-2)/u = Cx^2, (y/x-2)/(y/x) = Cx^2,
y-2x = Cyx^2, y(1) = 1 代入得 C = -1， 特解 y-2x = -yx^2

Answer 2

解:微分方程为x²y'+xy=y²，化为x²y'/y²+x/y=1，设1/y=u，有-x²u'+xu=1，-u'/x+u/x²=1/x³，(-u/x)'=1/x³，

-u/x=-0.5×1/x²+0.5c(c为任意常数)，2u=1/x-cx，方程的通解为2/y=(1-cx²)/x

∵y(1)=1 ∴得:c=-1，方程的特解为y=2x/(1+x²)

Answer 3

解:微分方程为x²y'+xy=y²，化为y'+y/x=(y/x)²，设y=ux，方程化为(ux)'+u=u²，u'x=u²-2u，du/[(u-2)u]=dx/x，2du/[(u-2)u]=2dx/x，ln|(u-2)/u|=lnx²+ln|c|(c为任意非零常数)，(u-2)/u=cx²，1-2/u=cx²，y-2x=cx²y ∵y(1)=1 ∴有c=-1，方程的特解为

y-2x+x²y=0