- #1

karush

Gold Member

MHB

- 3,269

- 5

$\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!