Find y2 given
t^2y'' + 3ty' + y = 0 y1(t) = 1/t
The Attempt at a Solution
i found y, y' and y'' and plugged it back in to the Differential equation,
after doing some work, i'm basicly down to
tv''+v' = 0
where y2 = v/t
so u = v' and u' = v'' Plugging that in...
tu' + u = 0
how would i solve this for u?
i think its separable equations but that would give me
du/-u = dt/t
which would be ln (u) = -ln(t) + c .... u = 1/t +c so v' = 1/t +c which means v = lnt
pluging v into y2 would give me lnt/t
Sorry its kind of confusing but the answer's not in the back of the book and i need to know if i did this right