Question

8t" alt=" \frac{dy}{dt} + 0.5yt=8t" align="absmiddle" class="latex-formula"> and $y(0)=9$

Find y(t)

Answer 1

$\dfrac{\mathrm dy}{\mathrm dt}+\dfrac t2y=8t$
$e^{t^2/2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac t2e^{t^2/2}y=8te^{t^2/2}$
$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t^2/2}y\right]=8te^{t^2/2}$
$e^{t^2/2}y=8\displaystyle\int te^{t^2/2}\,\mathrm dt$
$e^{t^2/2}y=8e^{t^2/2}+C$
$y=8+Ce^{-t^2/2}$

With $y(0)=9$, we get

$9=8+Ce^0\implies C=1$

and so the particular solution must be

$y=8+e^{-t^2/2}$