Question

全微分方程(x^2-y)dx+axdy=0(a为常数)的通解为

Answer 1

dy/dx=(x^2-y)/(-ax)

(x^2-y)/(-ax)=2u

y=x^2+2axu

dy=2xdx+2adu+2axdu

2x+2(a+ax)du/dx=2u

x-u= -(a+ax)du/dx

(x-u)/ -(a+ax)=du/dx

(x-u)/-(1+x)=v

u=x-v-xv

du=dx-dv-vdx-xdv

v/-a=(1-v)+(-x-1)dv/dx

(v/a-1+v)=(-x-1)dv/dx

-dx/(x+1)=dv/[v(1/a+1)-1)]

-ln(x+1)=[a/(1+a)]ln(v(1/a+1)-1]+C0

C1(x+1)^(-1/a-1)=v(1/a+1)-1

v=C2(x+1)^(-1/a-1)+a/(1+a)

u=x-v-xv

=x-C2(x+1)^(-1/a-1)-a/(1+a)-C2x^(-1/a)

y=x^2-2axu

=x^2-2ax^2+C3(x-1)^(-1/a)+[2a^2/(1+a)]x+C3x^(-1/a+1)