Question

y'+x=√(x^2+y)的微分方程的通解

Answer 1

y=x^2z
dy=2xzdx+x^2dz
2xzdx+x^2dz+xdx=x√(1+z)dx
2zdx+xdz+dx=√(1+z)dx
xdz=(√(1+z)-2z-1)dx
dz/(√(1+z)-2z-1)=dx/x
lnx=∫dz/(√(1+z)-2(z+1)+1)
=2∫√(1+z)d√(1+z)/[(2√(1+z)+1)(1-√(1+z))]
=(2/3)∫d√(1+z)/(1-√(1+z)) -(2/3)∫d√(z+1)/(2√(1+z)+1)
=(-2/3)ln|2√(1+z)+1|+(-2/3)ln|√(1+z)-1|+C
通解lnx=(-2/3)ln|2√(1+y/x^2)+1| +(-2/3)ln|√(1+y/x^2)-1|+C

√(1+z)/(√(1+z)-2(z+1)+1) =√(1+z)/(2√(1+z)+1)(-√(1+z)+1)
=(1/3)[(1/(-√1+z)+1)-1/(2√(1+z)+1)]

追问

可不可以用u=√(x^2+y)，化为齐次线性方程组

追答

u=√x^2+y
dy=2udu-2xdx
2udu-2xdx+xdx=udx
2udu=(u-x)dx
u=xv du=xdv+vdx
2xv(xdv+vdx)=(xv-x)dx
2vxdv+2v^2dx=(v-1)dx
2vxdv=(v-1-2v^2)dx
2vdv/(v-1-2v^2)=dx/x
lnx= -∫2vdv/(2v+1)(v-1)=(-1/3)ln(2v+1)+(-2/3)ln(v+1)+C
结果是相同的