Use reduction of order to find a second solution of the given differential equation(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Reduction of order

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