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)`.