The differential equation

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

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

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

$$

The reduced form has a common factor of $t^3$ which we can divide out of the equation. Since this equation does not have any $v$ terms in it we can make the substitution $u=v^{\text{'}}$ giving us the first order linear equation in $u$ :

$=0$

If we use c as the constant of integration, the solution to this equation is $u=$

Integrating to get $v,$ and then finding $y_2$ gives the general solution: