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

$

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- Jul 14th 2009, 12:56 PMdiddledabble2nd linearly independent solution using reduction of order
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

$ - Jul 14th 2009, 07:26 PMSampras
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.