$t^2y'' - 3ty' + 4y = 0$, how do you solve for $y$ in terms of $t$?

Question

I know that I have to use method of reduction, but I'm not sure how to arrive at a solution.

All help will be much appreciated.

Answer 1

trial solution $y = t^\alpha v$ $$ t^{\alpha+2}v'' + (2\alpha - 3)t^{\alpha+1}v' + \left[\alpha(\alpha-1)-2\alpha + 4\right]t^\alpha v = 0 $$ lets try get rid of the last term and hope all other terms are easier afterwards.. $$ \alpha(\alpha-1)-2\alpha + 4 = \alpha^2 - 4\alpha + 4 = 0\implies \alpha = 2 $$ now we have reduced to

$$ t^4v'' + t^3v' = 0 $$ which is $$ v'' + \frac{1}{t}v' = 0\tag{*} $$ we can reduce by setting $v' = u$ then $$ u' + \frac{1}{t}u =0 $$

$\textbf{edit}$

As in another Jacks solution..we can write Eq(*) as $$ tv'' + v' = (tv')' = 0 \implies tv' = C_1 $$

Answer 2

This is a Cauchy–Euler equation which is solved by the substitution $y=t^r$.
$$0=t^2r(r-1)r^{r-2}-3trt^{r-1}+4t^r=t^r(r^2-4r+4)=t^r(r-2)^2$$ As you have a root of multiplicty $2$, your solution will be of the form $$y=C_1t^2+C_2t^2\ln t$$

Answer 3

$$ t^2 y'' - 3t y' + 4\lambda y = 0 \tag{1}$$ is a Sturm-Liouville-type differential equation. To solve it, you can follow what is sketched in the paragraph Rodrigues' formula here, reaching $R(x)=x^3,W(x)=x$. When $\lambda=1$, by setting $y(t)=t^2 f(t)$ we get rid of the $y$ term in $(1)$, reaching: $$ t^3 (f'(t) + t f''(t)) = 0, $$ or just: $$ f'(t) + t\,f''(t) = 0, \tag{1}$$ from which it follows that: $$ f'(t) = \frac{K}{t},\qquad f(t)= C+K\log t,\qquad \color{red}{y(t) = Ct^2 + Kt^2 \log t}.\tag{3}$$