Question

Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t> 0 Given yı(t) = t y2(t) = Preview Give your answer in simplest form (ie no coefficients)

Answer 1

Let $y_2(t)=tv(t)$. Then

${y_2}'=tv'+v$

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

and substituting these into the ODE gives

$t^2(tv''+2v')+3t(tv'+v)-3tv=0$

$t^3v''+5t^2v'=0$

$tv''+5v'=0$

Let $u(t)=v'(t)$, so that $u'(t)=v''(t)$. Then the ODE is linear in $u$, with

$tu'+5u=0$

Multiply both sides by $t^4$, so that the left side can be condensed as the derivative of a product:

$t^5u'+5t^4u=(t^5u)'=0$

Integrating both sides and solving for $u(t)$ gives

$t^5u=C\implies u=Ct^{-5}$

Integrate again to solve for $v(t)$:

$v=C_1t^{-6}+C_2$

and finally, solve for $y_2(t)$ by multiplying both sides by $t$:

$tv=y_2=C_1t^{-5}+C_2t$

$y_1(t)=t$ already accounts for the $t$ term in this solution, so the other independent solution is $y_2(t)=t^{-5}$.