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Let consider ordinary differential equation of the form

$$t^2y''+3ty'+y=0$$

This is equivalent to

$$y''+\frac{3}{t}y'+\frac{1}{t^2}y = 0$$ which looks better. But how does one find the solutions here? I guessed one of them is $y(t)=\frac{1}{t}$, but guessing shouldn't be the method here. I feel as though I needed a smart substitution. Any hints?

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    $\begingroup$ Since $y_1(t) = \frac{1}{t}$ is a solution, you can write $y_2(t) = k(t) \frac{1}{t}$, and solve for $k(t)$, making $y_2$ a solution too. But I suppose that you don't want this. I'll see if I think of something. $\endgroup$ – Ivo Terek Sep 13 '14 at 19:49
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$$ t^2y'' + 3ty' + y = 0 $$ first thing to notice is that a trival (yet insightful(I believe) separation of terms) $$ t^2y'' + 2ty' + ty' + y = \dfrac{d}{dt}t^2y' + \dfrac{d}{dt}ty =0 $$

$$ \dfrac{d}{dt} \left(t^2y' + ty\right) = 0 $$ or $$ t^2y'+ty = C_1 $$ this is a first order ode which you can solve readily.

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    $\begingroup$ That is indeed insightful +1. $\endgroup$ – Jules Sep 13 '14 at 19:52
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    $\begingroup$ Very nice! Very well done indeed! +1! $\endgroup$ – Robert Lewis Sep 13 '14 at 20:05

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