$c=0$

$2.$ Solve the initial value problem: $t\frac{dy}{dt}=-y+2t,y(0)=0.$ Determine the t$-$interval in which the solution is defined. $(5$ pts $).$

$\frac{+dy}{dt}=\frac{-y+2t}{t},y(0)=0$

$\frac{dy}{dt}=\frac{-1}{t}y+2$

$x(t)=e^{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{t}}$

$x(t)=e^{ln(t)}=t$

$yt={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}2tdt+c$

$yt=t^2+C$

$y=t+\frac{c}{t}++\frac{d}{r}0$

$y=t+\frac{c}{t}$

$y(0)=0$

$0=0+\frac{c}{0}=$ underines $aty(0)$

theretore the t$-$interral is

$(-,0)(0,)$ because it cantrike $?$ It awever to make$)$ the f$-$inererval $($antinusar$,$ we choose $(0,)$