Question

微分方程x y'' = y' ln ( y' / x )

Answer 1

xy''=y'ln(y'/x)

x(y''/y')=ln(y'/x)

x(lny')'=lny'-lnx

lny'=p

xp'=p-lnx

xdp=pdx-lnxdx

p/x=u

dp=xdu+udx

x^2du+xudx=xudx-lnxdx

x^2du=-lnxdx

du=-lnxdx/x^2

u=∫-lnxdx/x^2=∫lnxd(1/x)=lnx/x-∫dx/x^2=lnx/x+1/x+C0

p=lnx+1+C0x

lny'=lnx+1+C0x

y'=Ce^(lnx+1+C0x)

y=∫Ce^(lnx+1+C0x)dx

=Ce∫xe^(C0x)dx

=(Ce/C0)∫xde^(C0x)

=(Ce/C0)xe^(C0x) - (Ce/C0)∫e^(C0x)dx

=(Ce/C0)xe^(C0x) - (Ce/C0^2)e^(C0x) +C