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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.

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

1 Answer 1

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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 $v$.

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