求特解xy'+y=yln(xy)求通解上边那求通解y''-ay'^2=0y(0)=0，y'(0)=-1求特解怎么没人答呢。

Question

求特解 xy'+y=yln(xy)求通解
上边那求通解
y''-ay'^2=0y(0)=0，y'(0)=-1求特解
怎么没人答呢。

Answer 1

xy' + y = yln(xy)
令t = xy,dt/dx = y + x•dy/dx
dy/dx = (1/x)(dt/dx - t/x)
x(1/x)(dt/dx - t/x) + t/x = (t/x)ln(t)
dt/dx = (t/x)ln(t)
dt/(tlnt) = (1/x) dx
d(lnt)/lnt = (1/x) dx
ln|lnt| = ln|x| + lnC₁ = ln|C₁x|
lnt = C₁x
t = e^(C₁x) = C₁e^x
xy = C₁e^x
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y'' - a(y')² = 0,y(0) = 0,y'(0) = -1
令t = dy/dx,dt/dx = d²y/dx²
dt/dx - a•t² = 0
dt/dx = a•t²
dt/t² = a•dx
-1/t = ax + C₁
t = -1/(ax + C₁),y'(0) = -1 t(0) = -1
-1 = -1/C₁ => C₁ = 1
dy/dx = -1/(ax + 1)
y = (-1/a)ln(ax + 1) + C₂,y(0) = 0,a≠0
0 = (-1/a)ln(0 + 1) + C₂ => C₂ = 0
∴特解为y = (-1/a)ln(ax + 1)