# solving 2nd order linear ODE with non constant coefficients

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?

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.
• 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. Apr 5, 2022 at 12:58
• 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. Apr 5, 2022 at 14:07
• I get almost the same, only the inverse. I think you missed a minus sign during the separation of variables transformation. $$\frac{z'}{z}=-\frac1t-\frac1{2(t+1)}, ~~~ u=\frac{z}{f^2}$$ Apr 5, 2022 at 15:05