Question

Use the method of reduction of order to find a second independent solution of the given differential equation. t2y'' + 3ty' + y = 0, t > 0; y1(t) = t−1

Answer 1

I assume you mean $y_1=t^{-1}$, and not $y_1=t-1$, since this doesn't satisfy the ODE.

Assume a second solution of the form $y_2=vy_1$, where $v$ is a function of $t$. Then

${y_2}'=v'y_1+v{y_1}'$

${y_2}''=v''y_1+2v'{y_1}'+v{y_1}''$

Substituting into the ODE gives

$t^2\left(\dfrac{v''}t-\dfrac{2v'}{t^2}+\dfrac{2v}{t^3}\right)+3t\left(\dfrac{v'}t-\dfrac v{t^2}\right)+\dfrac vt=0$

$\implies tv''+v'=0$

$\implies(tv')'=0$

$\implies tv'=C$

$\implies v'=\dfrac Ct$

$\implies v=C_1\ln|t|+C_2$

$\implies y_2=\dfrac{\ln t}t$

where we omit the second term because it's already accounted for by $y_1$.