- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

$\displaystyle y^\prime - \frac{2}{t}y

=\frac{\cos{t}}{t^2};

\quad y{(\pi)}=0, \quad t>0$

$u(t)=e^{2 \ln{t}}$

then

$\displaystyle e^{2\ln{t}}\, y^\prime - \frac{2e^{e^{2\ln{t}}}}{t}y

= \frac{e^{2\ln{t}}\cos{t}}{t^2}$

not sure actually!