Attempt $1$:$20$ attempts remaining.

The differential equation

$t^2{y}^{\text{'}\text{'}}-t(t2)y^{\text{'}}(t2)y=0$

has $y_1=t$ as a solution.

Applying reduction of order we set $y_2=v*y_1=v*t.$

Then $($using the prime notation for the derivatives$)$

$y_2^{\text{'}}=$

$y_2^{\text{'}\text{'}}=$

$$

$$

So$,$ substituting $y_2$ and its derivatives into the left side of the differential equation, and reducing, we get

$t^2{y}_2^{\text{'}\text{'}}-t(t2)y_2^{\text{'}}(t2)y_2=$

$$