tut0401 $y^{'} + \frac{2}{3}y=1-\frac{1}{2}t$, $y(0) = y_0$
$p(t) = \frac{2}{3}$, $g(t) = 1-\frac{1}{2}t$, then $u = e^{\int \frac{2}{3}dt}$,
multiply both sides with $u$, then we get:
$e^{\frac{2}{3}t}y^{'} + \frac{2}{3}e^{\frac{2}{3}t}y = e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t}$
$(e^{\frac{2}{3}t}y)^{'}=e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t}$
$d(e^{\frac{2}{3}t}y) = (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$
$e^{\frac{2}{3}t}y=\int (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$
$e^{\frac{2}{3}t}y=\frac{3}{2}e^{\frac{2}{3}t}-\frac{3}{4}te^{\frac{2}{3}t}+\frac{9}{8}e^{\frac{2}{3}t}+C$ (integrate by parts)
$y= \frac{3}{2}-\frac{3}{4}t+\frac{9}{8}+Ce^{-\frac{2}{3}t}$
$y=\frac{21}{8}-\frac{3}{4}t+Ce^{-\frac{2}{3}t}$. Consider $y(0) = y_0 \implies y_0=\frac{21}{8}-0+C \implies C=y_0-\frac{21}{8}$
Therefore, $y=\frac{21}{8}-\frac{3}{4}t+(y_0-\frac{21}{8})e^{-\frac{2}{3}t}$
If $y(t)$ touches, but does not cross the $t$-axis at some point $t_0$ s.t. $y(t_0) = 0$ and $y^{'}(t_0) = 0$
Thus, $y^{'} +\frac{2}{3}y = 1- \frac{1}{2}t$ when $t=t_0$
$0+0 = 1-\frac{1}{2}t_0$
$\frac{1}{2}t_0 = 1 \implies t_0 = 2$
substitute $t_0 = 2$, then $y(2) = 0 \implies 0 = \frac{21}{8} - \frac{3}{4}\times 2 + (y_0-\frac{21}{8})e^{-\frac{4}{3}}$
Therefore, we get $y_0 = \frac{21}{8}e^{-\frac{4}{3}}-\frac{9}{8}$
