# Second order ODE $y''+p(t)y'+q(t)y=0$

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?

• 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. – Ivo Terek Sep 13 '14 at 19:49

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