Answers: 2
is a known solution, we look for a solution of the form 
. It has derivatives 
and 
. Substituting these into the ODE gives
so that 
and we get an ODE linear in 
:
:![x^2e^{-x}w'+(2x-x^2)e^{-x}w=0/tex]Since [tex](x^2e^{-x})=(2x-x^2)e^{-x}](/tpl/images/0074/0239/a6e08.png)
, we can condense the left side as the derivative of a product:
:
. Unfortunately, there is no closed form for the integral of the right side, but we can leave the result in the form of a definite integral:
is any point on an interval over which a solution to the ODE exists.
to solve for 
:
term above, so the second independent solution is
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