Question

帮忙解一个微分方程:f^2(x) =2xf(x)/3 +x^2 f'(x)/3,且f(2)=2/9 ,求y=f(x)的表

Answer 1

令y=f(x)

y^2=2xy/3+x^2y'/3

3y^2=2xy+x^2y'

3(y/x)^2=2(y/x)+y'

令u=y/x y=ux y'=u'x+u

3u^2=3u+u'x

3u^2-3u=xdu/dx

dx/x=du/(3u^2-3u)

∫dx/x=∫du/3u(u-1)

ln|x|=1/3*(ln|u-1|-ln|u|)+C

x^3=C*(u-1)/u

因为f(2)=2/9

所以当x=2时,u=y/x=(2/9)/2=1/9

8=C*(1/9-1)/(1/9)

C=-1

所以x^3=1/u-1

u=1/(x^3+1)

y/x=1/(x^3+1)

y=x/(x^3+1)