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I am trying to solve the following 2nd order ODE with non constant coefficients.

$y''(t)+\frac{2+3t}{2t(t+1)}y'(t)-\frac{3}{2t(t+1)}y(t)=0$

Since I read that there is no general way of approaching to solve 2nd order linear ODE with non coefficients, how do I approach this problem?

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The last two terms are a multiple of $f(t)y'-f'(t)y$ with $f(t)=3t+2$ linear. This gives $y_1(t)=f(t)$ as one solution, as $f''(t)=0$.

Now try to find the other basis solution per reduction-of-order, that is, by setting $y_2(t)=f(t)u(t)$, $z(t)=f(t)^2u'(t)$, using $g(t)=2t(t+1)$, $$ 0=g[fu''+2f'u']+f[fu'],\\ 0=gz' + fz $$ This is now first-order linear, so solvable, in principle.

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  • $\begingroup$ Okay writing out your last equation, I substituted u''=w' and u'=w to reduce it to 1st order ODE. I solved this equation and now I have to integrate my solution w one more time to get u(t). In the end there should only be 1 integration constant left in the solution u(t) correct ? And I form my general solution as: y(t) = c1 f(t) + c2 f(t)u(t). The integration constant for u(t) goes into c2 I assume, so I am only left with 2 degrees of freedom for c1 and c2 ? $\endgroup$
    – trynerror
    Apr 5, 2022 at 12:18
  • $\begingroup$ You have 2 integrations, so you get 2 integration constants. But you can select them to be some convenient value, as you only want one solution as second basis solution. If you leave both constants undetermined, then $fu$ will already be the full general solution. $\endgroup$ Apr 5, 2022 at 12:58
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    $\begingroup$ I am a bit confused about your definitions and rewriting of the ODE/I dont get from your definitions back to the equation. Setting y=fu, y'=f'u+fu', y''=f''u+2f'u'. Plugging this into the ODE and setting g=(2t(t+1))^-1 I arrive at the equation fu''+(2f'+f^2g)u' = 0 and I cant map this to your result $\endgroup$
    – trynerror
    Apr 5, 2022 at 14:03
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    $\begingroup$ I left out the terms with factor $u$, as they combine to zero. Note that you took the inverse of my $g$ as your $g$. You can combine $fu''+2f'u'=\frac1f(f^2u')'$, this just combines some terms so that the remaining equation looks neater, it does not really change the solution process. $\endgroup$ Apr 5, 2022 at 14:07
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    $\begingroup$ I get almost the same, only the inverse. I think you missed a minus sign during the separation of variables transformation. $$\frac{z'}{z}=-\frac1t-\frac1{2(t+1)}, ~~~ u=\frac{z}{f^2}$$ $\endgroup$ Apr 5, 2022 at 15:05

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