Question

已知函数满足微分方程xy'=yln(yIx),且x=1时,Y=e^2是当x=-1时,Y=

Answer 1

xy' = yln(y/x)

令y = xv,y' = v + x · dv/dx = v + x · v'

v + x · v' = v · ln(v)

v' = (vln(v) - v)/x

∫ dv/[v(ln(v) - 1)] = ∫ 1/x dx

∫ d(ln(v) - 1)/[ln(v) - 1] = ∫ 1/x dx

ln[ln(v) - 1] = ln(x) + lnC = ln(Cx)

ln(v) = Cx + 1

v = e^(Cx + 1)

y/x = e^(Cx + 1)

y = xe^(Cx + 1)

当x = 1,y = e²

e² = e^(C + 1) => 2 = C + 1 => C = 1

∴通解y = xe^(x + 1)

当x = - 1,y = - 1 · e^(- 1 + 1)

= - 1