Differential Equations: Use method of reduction to find the second solution. t^2y''-2ty'-2y=0 y1(t)=t t0

Question

Differential Equations: Use method of reduction to find the second solution. t^2y''-2ty'-2y=0  y1(t)=t   t>0

anonymous · Answer

$$t^2y''-2ty'-2y=0\
y''-\frac{2}{t}y'-\frac{2}{t^2}y=0$$
Let $y_2(t)=v(t)y_1(t)=v(t)\cdot t,$ giving you
$$y_2'(t)=t~v'(t)+v(t) \
y_2''(t)=t~v''(t)+2v'(t)$$
Substituting into the original equation,
$$t~v''(t)+2v'(t)-\frac{2}{t}(t~v'(t)+v(t))-\frac{2}{t^2}t~v(t)=0\
t~v''(t)+2v'(t)-2v'(t)-\frac{2}{t}v(t)-\frac{2}{t}~v(t)=0\
t~v''(t)-\frac{4}{t}\color{red}{v(t)}=0$$
The $v(t)$ term should have disappeared (as is the case when doing reduction of order). I suspect you copied your diff.eq incorrectly.