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Consider the following second order differential equation:

$$ \frac{d^2y}{dt^2} + \frac{2v +1}{t} \frac{dy}{dt} + y = 0, v \in \mathbb{R} $$

If this has a solution $y(t)$ that is smooth on an interval about $0$. Show that $y(-t)$ is also a solution.

My obstacle is I do not know how to use smoothness of $y$ to get started.

Any hint is appreciated.

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so it means that v is constant right?i mean it is some number – dato datuashvili Mar 28 '13 at 6:44
it is also interesting it's solution,what would be characteristic equation of this model? – dato datuashvili Mar 28 '13 at 6:52
up vote 3 down vote accepted

Let $z(t) = y(\tau)$, where $\tau = -t$. We then have $$\dfrac{dz(t)}{dt} = \dfrac{dy(\tau)}{d \tau} \cdot \dfrac{d \tau}{dt} = - \dfrac{dy(\tau)}{d \tau} \,\,\, (\spadesuit)$$ $$\dfrac{d^2z(t)}{dt^2} = \dfrac{d}{dt}\left(- \dfrac{dy(\tau)}{d \tau} \right) = -\dfrac{d}{d \tau}\left(\dfrac{dy(\tau)}{d \tau}\right) \dfrac{d \tau}{dt} = \dfrac{d^2 y(\tau)}{d \tau^2} \,\,\, (\clubsuit)$$ We have that $y(\tau)$ satisfies $$\dfrac{d^2 y(\tau)}{d \tau^2} + \dfrac{2v+1}{\tau} \dfrac{d y(\tau)}{d \tau} + y(\tau) = 0$$ Making use of $(\spadesuit)$ and $(\clubsuit)$, we now get $$\dfrac{d^2 z(t)}{d t^2} + \dfrac{2v+1}{(-t)} \left(-\dfrac{d z(t)}{d t} \right) + z(t) = 0$$ which gives us $$\dfrac{d^2 z(t)}{d t^2} + \dfrac{2v+1}{t} \dfrac{d z(t)}{d t} + z(t) = 0$$

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