Ordinary differential equation homogeneous

I was solving ODE $$x^5yy'+y^4x^2-xy^5=0$$ with condition $$y\left(2\right)=1$$. Since it is homogeneous equation i substitute $$y=vx$$ and calculate furthur but i stuck at integration which is $$\int\dfrac{1}{v^4-v^3-v}dv.$$

$$\int\dfrac{1}{v^4-v^3-v}dv=\int \dfrac{1}{v\left(v^3-v^2-1\right)}dv.$$

Then i did partial fraction and i get $$\dfrac{-1}{v}+\dfrac{v^2-v}{v^3-v-1}$$. I don't know how to deal with second. (I am not sure i did correcr partial fraction).

Any hint how to integrate. Thank you.

• my mistake, i forgot to add y in first term Oct 6, 2020 at 7:56
• Your partial fraction is correct. Oct 6, 2020 at 8:01
• Does this help? wolframalpha.com/input/… Oct 6, 2020 at 8:04
• Yes it does thanks. Oct 6, 2020 at 8:09

You can do one easy reduction with one additional logarithmic derivative, $$(\ln|x|)'=\frac{v'}{v(v^3-v^2-1)}=-(\ln|v|)'+\frac13(\ln|y^3-y^2-1|)'-\frac13\frac{vv'}{v^3-v^2-1}.$$ For the last term, let $$a=1.46557123...$$ be the positive root of $$0=p(v)=v^3-v^2-1$$, then $$p(v)=(v-a)([v^2+av+a^2]-[a+v])=(v-a)(v^2+a^{-2}v+a^{-1})$$ The extended partial fraction decomposition relating to the last term then is $$\frac{v}{v^3-v^2-1}=\frac{v}{(v-a)(v^2+a^{-2}v+a^{-1}))}=\frac{a^2}{(a^2+3)(v-a)}+\frac{v - a^{-2}}{(2-3a)(v^2+a^{-2}v+a^{-1})}$$ which gives another two logarithmic terms and an inverse tangent term, as $$v^2+a^{-2}v+a^{-1}=(v+\tfrac12(a-1))^2+\tfrac14(3a+1)(a-1)$$ This all together is in the end not very well solvable for $$v$$.