- #1
Punchlinegirl
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Use reduction of order to find a second solution of the given differential equation
t^2y"-4ty'+6y=0 t>0 y_1(t)=t^2
first I put in in the standard form
y"-4/ty'+6/t^2y =0
then y= v(t)t^2
y'=v'(t)t^2+v(t)2t
y"=y"(t)t^2+v'(t)2t+v'(t)2t+v(t)2 = t^2v"(t)+4tv'(t)+2v(t)
putting these into the original equation gives me
t^2v"(t)+4tv'(t)+2v(t)-4tv'(t)-8v(t)+6t^2v(t)
simplifying gives me
t^2v"(t)+6t^2v(t)-6v(t)
I don't know what to do from here. can someone please help?
t^2y"-4ty'+6y=0 t>0 y_1(t)=t^2
first I put in in the standard form
y"-4/ty'+6/t^2y =0
then y= v(t)t^2
y'=v'(t)t^2+v(t)2t
y"=y"(t)t^2+v'(t)2t+v'(t)2t+v(t)2 = t^2v"(t)+4tv'(t)+2v(t)
putting these into the original equation gives me
t^2v"(t)+4tv'(t)+2v(t)-4tv'(t)-8v(t)+6t^2v(t)
simplifying gives me
t^2v"(t)+6t^2v(t)-6v(t)
I don't know what to do from here. can someone please help?