How do you find the second linearly independent solution using reduction of order.
Examples
$\displaystyle
t^{2}y'' + 6ty' +6y = 0
t > 0
f(t) = t^{-2}
$
$\displaystyle
ty'' + (1-2t)y' + (t-1)y = 0
t > 0
f(t) = e^t
$
How do you find the second linearly independent solution using reduction of order.
Examples
$\displaystyle
t^{2}y'' + 6ty' +6y = 0
t > 0
f(t) = t^{-2}
$
$\displaystyle
ty'' + (1-2t)y' + (t-1)y = 0
t > 0
f(t) = e^t
$
Suppose we have $\displaystyle t^{2}y'' + 6ty' +6y = 0, \ t > 0, \ y_{1}(t) = t^{-2}$.
We know that the second solution will have the form $\displaystyle y_{2}(t) = v(t)y_{1}(t) $. Now:
$\displaystyle y_{2}(t) = t^{-2}v $
$\displaystyle y_{2}'(t) = -2t^{-3}v+t^{-2}v' $
$\displaystyle y_{2}''(t) = \cdots $
Plug these into differential equation. You will get an expression only involving derivatives of $\displaystyle v $. Then make a substitution to get a linear first order differential equation.