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?
Best Answer
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.