Answer
$y_1(t) = t^{1/2}$ and $y_2(t) = t^{-1}$ are solutions to the differential equation $2t^{2}y''+3ty'-y = 0$, for $t>0$.
Work Step by Step
We must show that $y_1(t) = t^{1/2}$ and $y_2(t) = t^{-1}$ are solutions for $2t^{2}y''+3ty'-y = 0$, for $t>0$.
So, for $y_1(t) = t^{1/2}$ we need only substitute $y_1$ for $y$ in the equation and perform the indicated operations: I.e.,
we have
$y_1' = \frac{1}{2}t^{-1/2}$,
and
$y_1'' = -\frac{1}{4}t^{-3/2}$.
So.
$2t^{2}y_1''+3ty_1'-y_1$
$= 2t^{2}(-\frac{1}{4}t^{-3/2}) + 3t(\frac{1}{2}t^{-1/2}) - t^{1/2}$
$= -\frac{1}{2}(t^{1/2})+\frac{3}{2}t^{1/2}-t^{1/2}$
$= 0$, which was to be shown.
For $y_2(t) = t^{-1}$, we first calculate
$y_2'(t) = -t^{-2}$,
$y_2'' = 2t^{-3}$.
Then
$2t^{2}y_2''+3ty_2'-y_2$
$= 2t^{2}(2t^{-3}) + 3t(-t^{-2}) - t^{-1}$
$= 4t^{-1} - 3t^{-1} - t^{-1}$
$=0$, which was to be shown.
